Question

Given that equilibrium constant for the reaction $$2S{O_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2S{O_3}\left( g \right)$$ has a value of $$278$$ at a particular temperature. What is the value of equilibrium constant for the following reaction at the same temperature?
$$S{O_3}\left( g \right) \rightleftharpoons S{O_2}\left( g \right) + \dfrac{1}{2}{O_2}\left( g \right)$$
$(A)6.0 \times {10^{ - 2}}$
$(B)1.3 \times {10^{ - 5}}$
$(C)1.8 \times {10^{ - 3}}$
$(D)3.6 \times {10^{ - 5}}$

Answer 1

Hint: All the reactions generally reach an equilibrium point. Equilibrium is that point of time at which the rate of forward reaction is equals to the rate of backward reaction. The reaction quotient of the reaction at this equilibrium stage is known as equilibrium constant. We will see in the solution how equilibrium constant changes for different reaction equations and temperature.

Complete answer:
We know that equilibrium is a state in chemical reaction when the rate of forward reaction is equal to the rate of backward reaction and the reaction quotient calculated at equilibrium is known as equilibrium constant. Equilibrium constant is written as ${K_{eq}}$ .
It depends upon the reaction equation and temperature. So, we are given the equations at the same temperature and hence, temperature will not affect our result.
Now, for $$2S{O_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2S{O_3}\left( g \right)$$ , the equilibrium constant is:
$${K_{eq}} = 278$$
For $$S{O_3}\left( g \right) \rightleftharpoons S{O_2}\left( g \right) + \dfrac{1}{2}{O_2}\left( g \right)$$ , we will first see the changes in the equation. The new equation is half the previous equation. So, the new constant will be the square root of the previous one. Also the new equation is the reverse of the given equation, then the new equilibrium constant will be reciprocal of the given constant.
Therefore, the new equilibrium constant is given as:
$K_{eq}' = \dfrac{1}{{\sqrt {{K_{eq}}} }}$
$K_{eq}' = \dfrac{1}{{\sqrt {278} }}$
Therefore, the new equilibrium constant is:
$K_{eq}' = 6 \times {10^{ - 2}}$
Hence, the correct option is $(A)6.0 \times {10^{ - 2}}$ .

Note:
There are two types of equilibrium constants. One is equilibrium constant at constant concentration which is written as the ${K_c}$ and the other one is equilibrium constant at constant pressure which is written as ${K_p}$ . They both depend upon the temperature and the chemical equation.