The value of Kc for the reaction
2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g) is 1\times 10^{-4}. At a given time, the composition of reaction mixture is [HI] = 2 \times 10^{-5} mol, [H_{_{2}}] = 1 \times 10^{-5} mol and [I_{2}] = 1 \times 10^{-5} mol. In which direction will the reaction proceed?

Answers (1)

Now, as per the given information in the question
2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)

It has also been specified that K_{c} =1 \times 10^{-4}

Now, when we apply the law of mass action the Equilibrium constant in this given equation we get;

K_{c} =\frac{[H_{2}][I_{2}]}{[HI]^{2}}

It is also known to us that the reaction quotient, Q_{c}, specifies the relative ratio of products to reactants at a given point in time.

Q_{c} =\frac{[H_{2}][I_{2}]}{[HI]^{2}}

Q_{c} =\frac{(1\times10^{-5})(1\times10^{-5})}{(2\times10^{-4})}

Q_{c} =\frac{1}{4}=0.25

Here, Q_{c} >K_{c}

Therefore, the reaction will proceed in reverse direction.

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