Download ppt - Chapter 19 Spontaneous Change: Entropy and Free Energy Dr. Peter Warburton [email protected]

Chapter 19Spontaneous Change: Entropy and Free Energy

Dr. Peter [email protected]://www.chem.mun.ca/zcourses/1051.php

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Spontaneous processes

We have a general idea of what we consider spontaneous to mean:

A spontaneous process WILL OCCUR in a system WITHOUT any outside action being performed on the system.

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Ice melting above zero Celcius is spontatneous

Object falling to earth is

spontaneous

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Ice will melt above zero Celcius.

We don’t have to DO anything!

Objects will fall to earth.

We don’t have to DO anything!

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Since we DON’T have to DO anything for these spontaneous processes to occur it APPEARS that an overall energy change

from potential energy

to kinetic energy

IS SPONTANEOUS

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Non-spontaneous processes

We have a general idea of what we consider non-spontaneous to mean:

A non-spontaneous process WILL NOT OCCUR in a system UNTIL an outside action is performed on the system.

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Ice freezing above zero Celcius is

non-spontatneous

Object rising from earth is

non-spontaneous

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We can make water freeze above zero Celcius by

increasing the pressure.

We can make an object rise from the earth by picking it

up.

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Since we DO have to ACT for these non-spontaneous processes

to occur it APPEARS that an overall energy change from

kinetic energy

to potential energy

IS NON-SPONTANEOUS

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Chemistry and spontaneity

We know there are chemical processes that are spontaneous because we can put the chemical system together and reactants become products without us having to do anything.

H3O+ (aq) + OH- (aq) 2 H2O (l)

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Chemistry and non-spontaneity

We know there are chemical processes that are non-spontaneous because we can put the chemical system together and reactants DO NOT become products.

The system we put together stays like it isUNTIL WE CHANGE SOMETHING!

2 H2O (l) 2 H2 (g) + O2 (g)

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Spontaneous vs. non-spontaneous

It is obvious by the examples we’ve looked at that the opposite of every spontaneous process is a non-spontaneous process.

In chemical systems we’ve seen that if we put a chemical system together a reaction occurs until the system reaches equilibrium.

Whether the forward reaction or the reverse reaction dominates depends on which of the two reactions is spontaneous at those conditions!

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Equilibrium and spontaneity are

related!

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Spontaneity and energy

In our examples it APPEARED that the

spontaneous process ALWAYS takes a system to a

lower potential energy.

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Spontaneity and energy

If this were true all exothermic processes would be spontaneous and all endothermic processes would be non-spontaneous.

THIS ISN’T TRUE! H2O

NH4NO3 (s) NH4+ (aq) + NO3

- (aq)is spontaneous even though

H = +25.7 kJ

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Recall the First Law

The First Law of thermodynamics stated that the energy of an ISOLATED system is constant.What’s the largest ISOLATED system we can think of?

It’s the UNIVERSE!The energy of the universe is

constant!

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Recall the First Law

On the universal scale, there is no overall change in energy, and so lower energy CANNOT be the only requirement for spontaneity.

There must be something else as well!

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Further proof lower energy isn’t enough

If an ideal gas expands into a vacuum at a constant

temperature, then

no work is done

and

no heat is transferred

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No work done and no heat transferred means

NO OVERALL CHANGE in

energy

of the system

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This spontaneous process has

no overall change in energy!

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Entropy

Entropy (from Greek, meaning “in transformation”)

is a thermodynamic property that relates

the distribution of the total energy of the system

to the available energy levels of the particles.

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Entropy

A general way to envision entropy is

“differing ways to move”Consider mountain climbers on a mountain.

Two factors affect the distribution of mountain climbers on a mountain:

Total energy of all the climbers and how many places can you stop

on the mountain

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Consider hungry mountain climbers

Hungry mountain climbers have

little total energy amongst themselves to

climb a mountain, so most of them are near the

bottom, while some are distributed on the lower parts of the mountain. Few energy levels

can be reached!

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Consider well fed mountain climbers

Well-fed mountain climbers have

more total energy amongst themselves to

climb a mountain, so the climbers will be

more spread out on the whole mountain.

More energy levels can be reached!

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Temperature and total energy

The total energy shared by molecules is related to the temperature.

A given number of molecules at a low temperature (less total energy) have less “differing ways to move” than the same number of molecules at a

high temperature (more total energy).

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Higher mountain means more places to stop

If a well-fed mountain climber tries to climb Signal Hill, they will most likely reach the

top. They have only a few places to stop (levels) because Signal Hill is a small

mountain.The same well-fed mountain climber on

Mount Everest has a greater number of places to stop (levels) because it is a

larger mountain.

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Volume and energy levels

The number of levels of energy distribution of molecules is related to

the volume.A given number of molecules in a

small volume have less “differing ways to move” than the same

number of molecules in a larger volume.

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Entropy

The greater the number of “differing ways to move” molecules can take

amongst the available energy levels of a system

of a given state (defined by temperature and volume, and number of molecules),

the greater the entropy of the system.

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Expansion into vacuum

A gas expands into a vacuum because the

increased volume allows for a greater

number of “differing ways to move” for

the molecules, even if the temperature is

the same.

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Expansion into vacuum

That is, the entropy increases when the gas is allowed to expand into

a vacuum.

Entropy increase plays a role in spontaneity!

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Entropy is a state function

Entropy, S, is a state function like enthalpy or internal energy.

The entropy of a system DEPENDS ONLY on the current state

(n, T, V, etc.) of the system, and NOT how the system GOT TO BE

in that state.

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Boltzmann equation and entropy

More available energy levels when the size of a box increases – like expanding a gas into a vacuum

– ENTROPY INCREASES!

More energy levels are accessible when the temperature increases – ENTROPY INCREASES!

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Change in entropy is a state function

Because entropy is a state function, then change in entropy S is ALSO

a state function.

The difference in entropy between two states ONLY depends on the

entropy of the initial and final states, and NOT the path taken to get there.

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Hess’s Law

Recall Hess’s Law – as long as we get from the same initial state to the same

final state then H will be the same regardless of the steps we add together.

Change in entropy S will work exactly the same way! As long as we get from the same initial state to the same final state then S will be the same regardless of

the steps we add together.

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n, T, V help define the number of states (number of available energy levels) the system

can have. The many “different ways to move” of

molecules in a particular state are called microstates.

Hopefully it makes sense that more total states should automatically mean more total

microstates.

The number of microstates is often symbolized by W.

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Playing cards

Say we have a deck of 52 playing cards.

Choosing one playing card is a state.

If we choose the first card out of the pack, there are 52 microstates for this first state.

The second card (second state) we choose has 51 microstates, and so on.

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Overall there are

W = 52! 8 x 1067

possible distributions

(total microstates) for

52 playing cards!

Playing cards

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If we flip 52 coins (a coin is one state), with two possible microstates

(heads or tails) each, there are W = 252 = 4.5 x 1015

possible distributions. (total microstates)

A deck of 52 playing cards has greater entropy than 52 coins!

Playing cards and coin flips

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Ludwig Boltzmann formulated the relationship between the number of

microstates (W) and the entropy (S).

S = k ln W

The constant k is the Boltzmann constant which has a value equal to the gas constant R divided by Avagadro’s

number NA

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S = k ln W

where k = R / NA

k = (8.3145 JK-1mol-1) / (6.022 x 1023 mol-1)

k = 1.381 x 10-23 JK-1

We can see the units for entropy will be Joules per Kelvin (JK-1)

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Measuring entropy change

From the units for entropy (JK-1)

we get an idea of how we might measure entropy change S

It must involve some sort of energy change relative to the

temperature change!

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Measuring entropy change

S = qrev / T

The change in entropy is the heat involved in a

reversible process at a constant temperature.

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Heat IS NOT a state function

Since heat IS NOT a state function we need a reversible process to make it ACT LIKE a

state function.

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Reversible processes

In a reversible process a change in one direction is exactly equal and opposite

to the change we see if we do the change in the reverse direction.

In reality it is impossible to make a reversible process without making an

infinite number of infinitesimally small changes.

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Reversible processes

We can however imagine the process is done

reversibly and calculate the heat involved in it, so

we can calculate the reversible entropy

change that could be involved in a process.

Srev = qrev / T

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Endothermic increases in entropy

In these three processes the

molecules gain greater “differing ability to

move.”

The molecules occupy more available

microstates at the given temperature, and so the entropy increases in all three processes!

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Generally entropy increases when…

…we go from solid to liquid.…we go from solid or liquid to gas.…we increase the amount of gas in a reaction.…we increase the temperature.…we allow gas to expand against a vacuum.…we mix gases, liquids, or otherwise make solutions of most types.

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Problem

Predict whether entropy increases, decreases, or we’re uncertain for the following processes or reactions:

(g) Cl (g) H (aq)OH 2(l) OH 2 (aq) Cl 2 d)

(s) Ag 2 (s) ZnO(s) OAg 2 (s) Zn c)

(g) O (l) Hg 2 (s) HgO 2 b)

(g) OH 2 (s) S 3 (g) SO (g) SH 2 a)

22-iselectrolys

2-

2

2

222

Answers: a) decreases b) increases c) uncertain d) increases

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Evaluating entropy and entropy changes

Phase transitions – In phase transitions the heat change does occur reversibly, so we can use the formula

Srev = qrev / Tto calculate the entropy change. In this case the heat is the enthalpy of the phase transition and the temperature is the transition temperature

S = Htr / Ttr

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Phase transitions – For water going from ice (solid) to liquid, Hfus = 6.02 kJmol-1 at the melting point (transition temp.) of 273.15 K (0 C)

Sfus = Hfus / Tmp

Sfus = 6.02 kJmol-1 / 273.15 K

Sfus = 22.0 JK-1mol-1

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Phase transitions – For water going from liquid to gas, Hvap = 40.7 kJmol-1 at the boiling point (transition temp.) of 373.15 K (100 C)

Svap = Hvap / Tbp

Svap = 40.7 kJmol-1 / 373.15 K

Svap = 109 JK-1mol-1

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Problem

What is the standard molar entropy of vapourisation Svap for CCl2F2 if its boiling point is -29.79 C and Hvap = 20.2 kJmol-1?

Answer: Svap = 83.0 JK-1mol-1

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Problem

The entropy change for the transition from solid rhombic sulphur to solid monoclinic sulphur at 95.5 C is Str = 1.09 JK-1mol-1. What is the standard molar enthalpy change Htr for this transition?

Answer: Htr = 402 Jmol-1

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Absolute entropies

Say we imagine a system of molecules that has no total energy. At this zero-point energy there can ONLY be ONE

possible distribution of microstates, as no molecule has the energy to occupy a

higher energy level.

The entropy CAN NEVER get smaller than its value in this situation, so we define

the entropy S of this situation as ZERO.

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Absolute entropies

This kind of imagining is the Third Law of Thermodynamics which states that

The entropy of a pure perfect crystal at 0K is zero.

At conditions other than at absolute zero, our entropy is that of the perfect system (zero)

PLUS any entropy changes that come changing temperature and/or volume.

These are absolute entropies!

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Methyl chloride entropy as a function

of temperature

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Standard molar entropies

One mole of a substance in its standard state will have an absolute entropy that we often call the standard molar entropy S.

These are usually tabulated at 298.15 K

In a chemical process we can then use these standard molar entropies to calculate the entropy change in the process.

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S = [pS(products) - rS(reactants)]

Hopefully this looks somewhat familiar!

We have seen a special treatment of Hess’s Law in Chem 1050 where

H = [pHf(products) - rHf(reactants)]

We can do something similar with ANY thermodynamic property that IS A STATE

FUNCTION!

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S = [pS(products) –

rS(reactants)]

H = [pHf(products) –

rHf(reactants)]

Enthalpies of formation ARE NOT absolute!

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Problem

Use the data given to calculate the standard molar entropy change for the synthesis of ammonia from its elements.

N2 (g) + 3 H2 (g) 2 NH3 (g)

S298 for N2 = 191.6 JK-1mol-1

S298 for H2 = 130.7 JK-1mol-1

S298 for NH3 = 192.5 JK-1mol-1Answer: -198.7 JK-1 mol-1 (per mole of rxn)

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Problem

N2O3 is an unstable oxide that readily decomposes. The decomposition of 1.00 mol N2O3 to nitrogen monoxide and nitrogen dioxide at 25 C is accompanied by the entropy change S = 138.5 JK-1mol-1. What is the standard molar entropy of N2O3 (g) at 25 C?

S298 for NO (g) = 210.8 JK-1mol-1

S298 for NO2 (g) = 240.1 JK-1mol-1

Answer: 312.4 JK-1 mol-1

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The second law of thermodynamics

We’ve seen that entropy MUST play a role in spontaneity,

because the total energy of the universe doesn’t change.

We could say that an entropy increase leads to spontaneity,

but we have to be careful.

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Ice freezing below 0 Celcius is spontaneous, but the entropy of the

water decreases in the process!

Sfreeze = -Hfus / Tmp

Since Hfus and Tmp are +ve,

then Sfreeze is –ve!-ve since “freezing” is the reverse of “fusion” (like we do in Hess’s Law)

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The water is ONLY the system.

The rest of the universe (the surroundings) must experience an

opposite heat change as it takes the heat the freezing water gave off (a +ve H for

the surroundings), which means

the entropy of the REST OF THE UNIVERSE INCREASES in the process

of water freezing!

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The total entropy change in any process is the entropy change for the system

PLUSthe entropy change for the surroundings

Suniverse = Stotal = Ssys + Ssurr

Now we can connect entropy and spontaneity!

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In any spontaneous process

the entropy of the universe INCREASES.

Suniverse = Ssys + Ssurr > 0

This is the Second Law of Thermodynamics!

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Water freezing

So while water freezing below zero Celcius decreases the entropy of the

system, the heat given off to the surroundings increases the entropy of the surroundings to a greater extent.

The total entropy change of the universe is positive and the process of

water freezing below 0 Celcius is spontaneous!

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Free energy

We’ve seen entropy increases when molecules

have more ways to distribute themselves amongst the

energy levels.

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Free energy

However, some of the energy a molecule uses to put itself at a higher energy level CAN NO LONGER be used to to do work

because doing work would put the molecule back at a lower energy level,

which would automatically decrease entropy.

The energy is NOT free (or available) to be used!

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S = Hrev / TReversible since the rest of the universe is SO BIG

then

TS = H

Free energy

Suniverse = Ssys + Ssurr

TSuniverse = TSsys + TSsurr

TSuniverse = TSsys + Hsurr

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Free energy

TSuniverse = TSsys + Hsurr

TSuniverse = TSsys - Hsys

Since Hsys = -Hsurr

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Free energy

TSuniverse = TSsys – Hsys-TSuniverse = Hsys - TSsys

G = Hsys – TSsys

G is the free energy(Gibbs free energy)

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Free energy

G = -TSuniv

For a spontaneous process Suniv > 0, which means for a spontaneous

process G < 0!

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Free energy

G < 0 is spontaneous

G > 0 is non-spontaneous

G = 0 is at equilibrium

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Problem

Predict the spontaneity at low and high temperatures for:

N2 (g) + 3 H2 (g) 2 NH3 (g) H = -92.22 kJ

2 C (s) + 2 H2 (g) C2H4 (g) H = 52.26 kJ

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Problem

N2 (g) + 3 H2 (g) 2 NH3 (g) H = -92.22 kJ

Spontaneous @ low T and nonspontaneous @ high T

2 C (s) + 2 H2 (g) C2H4 (g) H = 52.26 kJ

Nonspontaneous @ all T

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Standard free energy change G

Just like we can have a standard enthalpy change H for chemicals, we can also define the standard free energy change…

G = H - TS

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Standard free energy of formation Gf

Standard enthalpies of formation Hf of elements in their standard states are zero:

Hrxn = [p Hf(products) –

r Hf(reactants)]Standard free energies of formation Gf of elements in their standard states are zero:

G = [p Gf(products) –

r Gf(reactants)]

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Problem

Calculate G at 298.15 K for the reaction

4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)by using the two following sets of data. Compare your answers.a) H = -1648 kJ mol-1

and S = -549.3 JK-1 mol-1

b) Gf (Fe) = 0 kJmol-1

Gf (O2) = 0 kJmol-1

Gf (Fe2O3) = -742.2 kJmol-1

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Problem answer

a) G = -1484 kJmol-1

b) G = -1484.4 kJmol-1

The answers are the same because free energy is a state function.

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Free energy and equilibrium

We’ve already seen that

G < 0 is spontaneous G > 0 is non-spontaneous

One process in a written reaction is spontaneous while the reverse is not, as long as G ≠ 0. Therefore

G = 0 is at equilibrium

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Water and steam

H2O (l, 1 atm) H2O (g, 1 atm)

G373.15 K = 0 kJmol-1

The system is at equilibrium at 1 atm (standard

conditions) and at the boiling point temperature!

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Water and steam


G298.15 K = 8.590 kJmol-1

The system is not at equilibrium at 1 atm

(standard conditions) and at the room temperature!

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Water and steam


G298.15 K = 8.590 kJmol-1

The forward process is non-spontaneous (G > 0) so the

reverse process is spontaneous and condensation occurs.

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Water and steam

H2O (l, 0.03126 atm)

H2O (g, 0.03126 atm) G298.15 K = 0 kJ

The system is at equilibrium at 0.03126 atm (non-standard

conditions) and at the room temperature!

87

Water and steam

H2O (l, 0.03126 atm)

H2O (g, 0.03126 atm) G298.15 K = 0 kJmol-1

Water CAN evaporate at room temperature,

just not to give an equilibrium pressure of 1 atm!

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Non-standard conditions

As we’ve seen with the previous water example,

our interest in an equilibrium system is often at non-standard conditions,

so knowing G is usually not as useful as

knowingG.

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Non-standard conditions

For an ideal gas H does not change if pressure changes, so at all non-

standard conditions H = H.

For an ideal gas S does change if pressure changes (expansion into vacuum shows us this!), so at all

non-standard conditions S ≠ S.

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Non-standard free energy

Because of these facts, the non-standard free energy change is

G = H - TSBut the standard free energy change

is

G = H - TS

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Non-standard free energy

The difference between standard and non-standard free energy is totally

due to the difference in entropy change between the standard and

non-standard conditions

G - G = - T(S-S)G = G + T(S-S)

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Boltzmann distribution

By the Boltzmann distribution

S = R ln W for one mole of particles

G = G + T(R ln W - R ln W)G = G + T(R ln W / W)

We are comparing a real system with a standard one. We are

dealing with activities!

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G = G + RT ln Qeq

In the comparison, we are looking at a reaction quotient!

In other words, we are looking at how our non-standard system is

different from the system we have at standard conditions!

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G = G + RT ln Qeq

If our system

is at equilibrium

then Qeq = Keq

and G = 0

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G = G + RT ln Keq = 0

which means

G = -RT ln Keq

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Equilibrium constant

The thermodynamic equilibrium constant for a reaction is directly

related to the standard free energy change!

G = -RT ln Keq

Keq = e-G/RT

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Equilibrium constant

G 0 Keq 1

G >> 0 then Keq is very small

G << 0 then Keq is very large

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Equilibrium constant at 298 K

G Keq Meaning

Keq = e-G/RT

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Predicting reaction direction

G < 0 means the forward reaction is spontaneous at the given non-standard conditions

G < 0 means the forward reaction is spontaneous at standard conditions and so K > 1

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G = 0 means the system is at equilibrium at the given non-standard conditions

G = 0 means the system is at equilibrium at standard conditions and so Keq = 1 which only occurs at one specific T!

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G > 0 means the forward reaction is non-spontaneous at the given non-standard conditions

G > 0 means the forward reaction is non-spontaneous at standard conditions and so K < 1

102


G = G

ONLY at standard conditions!

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Thermodynamic equilibrium constant Keq

The thermodynamic equilibrium constant Keq is expressed in terms of activities, which

are unitless quantities.

Activities relate properties like concentration or pressure compared to a standard property

value, like 1 M for concentration or 1 bar for pressure.

104

Thermodynamic equilibrium constant Keq

a A + b B c C + d D

b

Ba

A

dD

cC

eqaa

aaK

where ax = [X] / c0 (c0 is a standard concentration of 1 M)

or ax = Px / P0 (P0 is a standard pressure of 1 atm)

Note that ax = 1 for pure solids and liquids

105

Free energy and equilibrium constants

G = -RT ln Keq

IS ALWAYS TRUEG = -RT ln Kc

andG = -RT ln Kp

DO NOT have to be true!

106

Problem

Write thermodynamic equilibrium constant expressions for each of the following reactions and relate them to Kc and Kp where appropriate:a) Si (s) + 2 Cl2 (g) SiCl4 (g)

b) Cl2 (g) + H2O (l) HOCl (aq) + H+ (aq) + Cl- (aq)

107

Problem answer

a) Si (s) + 2 Cl2 (g) SiCl4 (g)

b) Cl2 (g) + H2O (l) HOCl (aq) + H+ (aq) + Cl- (aq)

2222

-

Cl

-

Cl

-

OHCl

ClHHOCleq P

]][Cl[HOCl][H

1P

P

c][Cl

c][H

c[HOCl]

aa

aaaK

p2

Cl

SiCl

2

Cl

SiCl

2ClSi

SiCleq K

P

P

PP

1

PP

aa

aK

2

4

2

4

2

4

108

Problem

Use the given data to determine if the following reaction is spontaneous at standard conditions at 298.15 K:

N2O4 (g) 2 NO2 (g)

Gf (N2O4) = 97.89 kJmol-1

Gf (NO2) = 51.31 kJmol-1

Answer: G = 4.73 kJmol-1, not spontaneous at standard conditions.

109

Problem

Based on the problem of the previous slide, determine which direction the reaction will go in if 0.5 bar of each gas is placed in an evacuated container:

N2O4 (g) 2 NO2 (g)

Answer: Since G = 4.73 kJmol-1, then Keq = 0.15 = Kp. Since Qp = 0.5 the reaction should go from right to left.

110

Problem

Determine the equilibrium constant at 298.15 K for the following reaction using the given data:

AgI (s) Ag+ (aq) + I- (aq)

Gf (AgI) = -66.19 kJmol-1

Gf (Ag+) = 77.11 kJmol-1

Gf (I-) = -51.57 kJmol-1

111

Problem answer

Since G = 91.73 kJmol-1, then Keq = 8.5 x 10-17 = Kc = Ksp.If we compare to the Ksp value in Table 18.1 (8.5 x 10-17) we see we are definitely in the right ballpark!

112

G and Keq are functions of T

We’ve already seen that equilibrium constants change with temperature.

Why?

Keq = e-G/RT

and

G = H - TS

113


G = -RT ln Keq = H - TS

ln Keq= -(H/RT) + (S/R)

114

Problem

At what temperature will the following reaction have Kp = 1.50 x 102?

2 NO (g) + O2 (g) 2 NO2 (g)

H = -114.1 kJmol-1

and S = -146.5 J K-1mol-1

Answer: T = 606 K

115


G = -RT ln Keq = H - TS

ln Keq= -(H/RT) + (S/R)

If H and S are constant over a temperature range

then

116


ln K1= -(H/RT1) + (S/R)minus

ln K2= -(H/RT2) + (S/R)

ln K1 - ln K2 =

-(H/RT1) - (-H/RT2)

117


ln K1 - ln K2 =

[-H/R] [(1T1) - (1T2)]OR

ln [K1/K2] =

[-H/R] [(1T1) - (1T2)]

118

van’t Hoff equation

The van’t Hoff equation relates equilibrium constants to

temperatures

ln [K1/K2] = [-H/R] [(1T1) - (1T2)]It looks very similar to the Arrhenius

equation and so a plot of ln K versus 1/T should give a straight line

with a slope of [-H/R]

119

Plot of ln Kp versus 1/T for the

reaction.

slope = -H / R, so for this reaction

H = -180 kJmol-1

2 SO2 (g) + O2 (g) 2 SO3 (g)

120

Problem

The following equilibrium constant data have been determined for the reaction

H2 (g) + I2 (g) 2 HI (g)

Kp = 50.0 @ 448 CKp = 66.9 @ 350 C

Estimate H for the reaction.

Answer: H = -11.1 kJmol-1.

121

Coupled reactions

If we have a non-spontaneous (G > 0) reaction with a product that appears as a reactant in a different reaction that is spontaneous (G < 0) then we can couple the two reactions (do them in the same container at the same time) to drive the non-spontaneous reaction!

122

Coupled reactions

This should make sense because if we have two equilibria in one container, one with a small K (like Ksp) and one with a large K (like Kf for complex formation), then the coupled reaction that is the sum of the two reactions has a K value that is larger than Ksp!

We have made the non-spontaneous reaction occur!