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# Reversible reaction

• Which statement is correct about a reaction for which the equilibrium constant is independent of temperature ?

A The requilibrium constant for the both forward and reverse reaction do not vary with temperature.

B The enthalpy change is zero.

C  The activation energies for both forward and reverse reactions are zero.

D  There are equal number of moles of reactants and products.

Why A is not correct here? increasing or decreasing the temperature will not aafect the value of Kc.

• Which statement is correct about a reaction for which the equilibrium constant is independent of temperature

A  the equlibrium constant for both the forwar and reverse raection do not vary with temperature.

B  the enthalpy change is zero.

C the activation energies for both forward and backward raction are zero

D there are equal number of moles of reactants and products.

why A is wrong? I have replaced rate constant with equlilibrium constant ....in this case choice A will still be wrong?

• Originally posted by hoay:

Which statement is correct about a reaction for which the equilibrium constant is independent of temperature

A  the equlibrium constant for both the forwar and reverse raection do not vary with temperature.

B  the enthalpy change is zero.

C the activation energies for both forward and backward raction are zero

D there are equal number of moles of reactants and products.

why A is wrong? I have replaced rate constant with equlilibrium constant ....in this case choice A will still be wrong?

For your modified question, both A and B are correct. For the original Cambridge question, only B is correct. Even if a reaction is neither endothermic or exothermic, increasing temperature will increase both the k-forward and k-backward rate constants, but the magnitude of increase will be equal, meaning that the Kc-forward and Kc-backward values will be unaffected.
• So Rate constant and equilibrium constant are different here for this case.

Can you present an exmaple of this type of raection?

• Originally posted by hoay:

So Rate constant and equilibrium constant are different here for this case.

Can you present an exmaple of this type of raection?

Cambridge was just testing the conceptual understanding of students, *IF* such a reaction exists.

In practice, for enthalpy change of a chemical reaction to be zero, is theoretically possible but extremely rare (in contrast, after taking entropy change into account, for Gibbs free energy change to be zero, ie. neither endogernic nor exogernic, at a particular temperature, is extremely common, eg. freezing / melting of water at 0 degrees C).

One such possible reaction (and in fact the only case within the A level syllabus), for enthalpy (and in this case, also entropy and Gibbs free energy) change of reaction to be zero, would be the conversion of 1 enantiomer to the other, eg. an anion as nucleophile replacing the same anion as leaving group via SN2, or via SN1 (which cheats a little by using overall reaction rather than individual elementary step-wise reactions).

There are other rare examples, but which are beyond A levels and need not be considered, and will not be asked by Cambridge.

• For a gaseous system...increasing pressure increases the rate of forward reaction as there are more colliding particles but why does it increases the rate for a reverse raection..do we assume that the increase in pressure also increases the no of particles of product side also...in case of 2nd stage of Contact process.

• Originally posted by hoay:

For a gaseous system...increasing pressure increases the rate of forward reaction as there are more colliding particles but why does it increases the rate for a reverse raection..do we assume that the increase in pressure also increases the no of particles of product side also...in case of 2nd stage of Contact process.

Increasing pressure of a system by decreasing its volume, increases the molarities of both reactants and products, hence increasing the rates of both forward and backward reactions.

But if the no. of moles of gaseous reactants vs products are different, the magnitude of increase in rates of forward vs backward reactions will be different (this can be shown mathematically as the stoichiometries of the reaction determine the Kc and Qc formulae), hence position of equilibrium will concordantly and consequently shift towards the side with less moles of gas (if pressure is increased by decreasing volume).