Question

${{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}:\text{ }{{\text{K}}_{c}}=1.2$. At the start of the reaction, there are 0.249mol of ${{N}_{2}}$, $3.21\times {{10}^{-2}}mol\text{ of }{{\text{H}}_{2}}\text{ and 6}\text{.42}\times \text{1}{{\text{0}}^{-4}}mol\text{ of N}{{\text{H}}_{3}}$ in a 3.5L reaction vessel at ${{375}^{\circ }}C$. Hence, the reaction will proceed in?
[A] Forward direction
[B] Backward direction
[C] Equilibrium
[D] Stops

Answer 1

Hint: To solve this question firstly find out the concentration of nitrogen, hydrogen and ammonia using the number of moles given and the volume. Use it to find out the reaction quotient by dividing the concentration of products by that of the reaction. Do not forget to include stoichiometry which calculates reaction quotient.

Complete step by step answer:
 In the question, the equilibrium constant is given to us and we have to find out whether the reaction will move forward, backward, attain equilibrium or will stop.
We know that the reaction quotient will give us the direction of the reaction. We denote the reaction quotient as ‘Q’.
Now, we know that the reaction quotient is given to us by the concentration of the products divided by that of the reactants.
The reaction given to us is ${{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}$
Therefore, we can write that Q = $\dfrac{{{\left[ N{{H}_{3}} \right]}^{2}}}{\left[ {{N}_{2}} \right]{{\left[ {{H}_{2}} \right]}^{3}}}$

The number of moles of the reactants and the product and the volume of solution is given to us. To calculate the concentration we can divide the number of moles by the volume.
Now, $\left[ N{{H}_{3}} \right]=\dfrac{\text{ 6}\text{.42}\times \text{1}{{\text{0}}^{-4}}mol\text{ }}{3.5L}=1.8343\times \text{1}{{\text{0}}^{-4}}mol/L$
$\left[ {{H}_{2}} \right]=\dfrac{\text{ }3.21\times {{10}^{-2}}mol\text{ }}{3.5L}=9.1714\times \text{1}{{\text{0}}^{-3}}mol/L$
$\left[ {{N}_{2}} \right]=\dfrac{\text{ 0}\text{.249}mol\text{ }}{3.5L}=0.0711\text{ }mol/L$

So putting the concentration values in the reaction quotient equation we will get-
     \[Q=\dfrac{{{\left( 1.8343\times \text{1}{{\text{0}}^{-4}} \right)}^{2}}}{\left( 0.0711 \right){{\left( 9.1714\times \text{1}{{\text{0}}^{-3}} \right)}^{3}}}=0.6134\]

The equilibrium constant is given to us as 1.2 and we have calculated the reaction quotient which came out to be 0.6134. Since, the reaction quotient is lower than equilibrium constant, therefore the reaction will move in the forward direction.
So, the correct answer is “Option A”.

Note: We should remember that-
- If the reaction quotient is higher than the equilibrium constant, the reaction will proceed in the backward direction.
- If the reaction quotient is lower than the equilibrium constant, the reaction proceeds in the forward direction.
- If the reaction quotient is equal to the equilibrium constant, then the system attains equilibrium and the reaction will not proceed in any direction.