Question

If the equilibrium constant for reaction (i) is 2, what is the equilibrium constant for reaction (ii)
$CO_{ 2 }\rightleftharpoons CO+1/2O_{ 2 }$ ……(i)
$2CO+O_{ 2 }\rightleftharpoons 2CO_{ 2 }$ ……(ii)
(A.) 1/4
(B.) 1/2
(C.) 1
(D.) 2

Answer 1

Hint: To answer this question you should know that if we multiply equation (i) with 2 and then invert it, we will get equation (ii). Now try to figure out the relation between the rate constants (equilibrium constant) of (i) and (ii).

Complete step by step answer:

Let’s find the correct answer to this question.
We will assume that the rate constant for the reaction (i) $CO_{ 2 }\rightleftharpoons CO+1/2O_{ 2 }$, is ${ K }_{ 1 }$ and for the reaction (ii) $2CO+O_{ 2 }\rightleftharpoons 2CO_{ 2 }$ is ${ K }_{ 2 }$.
First, we have multiplied the reaction (i) with 2,
$2CO_{ 2 }\rightleftharpoons 2CO+O_{ 2 }$
so, the new rate constant (suppose K) will be equal to ${ K }_{ 1 }^{ 2 }$.
Now, we will invert the reaction above to get a reaction identical to reaction (ii)
$2CO+O_{ 2 }\rightleftharpoons 2CO_{ 2 }$
Here, the final rate constant, ${ K }_{ 2 }$ will be equal to 1/${ K }_{ 1 }^{ 2 }$.
We already have the value of ${ K }_{ 1 }$ = 2 in the question.
So, ${ K }_{ 2 }$ = 1/${ 2 }^{ 2 }$ = 1/4

Therefore, we can conclude that the correct answer to this question is option A.

Note: Here we should know that the so-called rate constant - which is only actually constant if all you are changing is the concentration of the reactants.
If you change the temperature or the catalyst, for example, the rate constant changes. This is shown mathematically in the Arrhenius equation.