Question

Calculate the value of $Q_c$ value, for the given reaction below. Also, given the concentration of all the species involved at a certain time:
$CO\left(g\right)+H_{2}O\left(g\right)\rightleftharpoons C{O}_2\left(g\right)+H_2\left(g\right)\left[C{O}_2\right]=2M\left[H_2\right]=2M\left[CO\right]=1M\left[H_{2}O\right]=1M$

If $K_c=1$ , in which direction the reaction will shift to attain equilibrium?

• A. $Q_C=2$ and forward
• B. $Q_C=2$ and backward
• C. $Q_C=4$ and forward
• D. $Q_C=4$ and backward

Answer 1

The correct option is D $Q_C=4$ and backward

We know for a reaction:
$aA\left(g\right)+bB\left(g\right)\rightleftharpoons cC\left(g\right)+dD\left(g\right)$
In terms of Concentration :
$Q_c=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$

So, In this case,
$Q_c=\frac{\left[C{O}_2\right].\left[H_2\right]}{\left[CO\right].\left[H_{2}O\right]}=\frac{2\times 2}{1\times 1}=4$
Comparing $Q_{c}and{K}_c$
Since, $Q_c>K_c$ the reaction shifts backwards (to the reactants side) for attaining equilibrium.

Theory:
Reaction quotient :
The reaction quotient (Q) measures the relative amounts of products and reactants present during a reaction at a particular point in time.
Reaction :
$aA\left(g\right)+bB\left(g\right)\rightleftharpoons cC\left(g\right)+dD\left(g\right)$
In terms of Partial Pressure :
$Q_p=\frac{\left[P{]}_c^c\right[P{]}_d^d}{\left[P{]}_a^a\right[P{]}_b^b}$
In terms of Concentration :
$Q_c=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$

Applications of Reaction quotient :
Case 1 :-
$Q_c>K_c$
For the given reaction :
$R\rightleftharpoons P$

If $Q_c>K_c$ or if $Q_c\ne K_c$ this means that system is not in equilibrium.
$Q_c$ will move close toward $K_c$ until it becomes equal to $K_c$.
Value of $K_c$(equilibrium constant) remains constant.
It promotes backward reaction when $Q_c>K_c$ to attain equilibrium.

Case 2 :- $Q_c<K_c$
For the given reaction :
$R\rightleftharpoons P$

If $Q_c<K_c$ or if $Q_c\ne K_c$ this means that system is not in equilibrium.
$Q_c$ will move close toward $K_c$ until it becomes equal to $K_c$. It promotes forward reaction when $Q_c<K_c$ to attain equilibrium.