Question

A mixture of 1.57 moles of ${ N }_{ 2 } $, 1.92 moles of ${ H }_{ 2 }$ and 8.13 moles of ${N H }_{ 3 }$ is introduced into a 20 L reaction vessel at 500 K. At this temperature the equilibrium constant ${ K }_{ c }$ for the reaction ${ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2{ NH }_{ 3 }(g)$ is $1.7\times { 10 }^{ 2 }$. What is the direction of the net reaction?
(a) Forward
 (b) Backward
(c) At equilibrium
(d) Data is inefficient

Answer 1

Hint: In order to determine the stage at which the given reaction is we need to find the reaction quotient for the reaction. If its value is equal to the equilibrium constant then the reaction is already at equilibrium. If its value is greater than the equilibrium constant then the reaction will go in the backward direction and if its value is less than the equilibrium constant then the reaction will go in the forward direction.

Complete Solution :
To determine at what stage the given reaction is, we need to find its reaction quotient. If the value of the reaction quotient comes out to be equal to equilibrium constant then it would mean that the reaction is already at equilibrium. If the value of the reaction quotient comes out to be greater than the equilibrium constant, then the reaction will go in the backward direction. If the value of the reaction quotient comes out to be less than the equilibrium constant then the reaction will go in the forward direction until the equilibrium is attained. Just like the equilibrium constant, the reaction quotient also depends upon the temperature.
${ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2{ NH }_{ 3 }(g)$
For the above reaction, the reaction quotient is: ${ Q }_{ c }=\cfrac { \left[ { NH }_{ 3 } \right] ^{ 2 } }{ \left[ { H }_{ 2 } \right] ^{ 3 }\left[ { N }_{ 2 } \right] } $

Now the concentration of ${ NH }_{ 3 } $ will be: $\left[ { NH }_{ 3 } \right] =\cfrac { 8.13\quad mol }{ 20\quad L } =0.4065\quad M$
$\left[ { N }_{ 2 } \right] =\cfrac { 1.57\quad mol }{ 20\quad L } =0.0785\quad M$
$\left[ { H }_{ 2 } \right] =\cfrac { 1.92\quad mol }{ 20\quad L } =0.096\quad M$

Therefore,
${ Q }_{ c }=\cfrac { (0.4065{ ) }^{ 2 } }{ (0.096{ ) }^{ 3 }(0.0785) } =2.379\times { 10 }^{ 3 }$
${ K }_{ c }=1.7\times { 10 }^{ 2 }$
Hence ${ K }_{ c }$ is not equal to ${ Q }_{ c }$ therefore the reaction is not at equilibrium.
The value of ${ Q }_{ c }>{ K }_{ c }$, therefore the reaction should go in the backward direction.
So, the correct answer is “Option B”.

Note: The reaction quotient is also called concentration quotient. If we find the ratio of the concentrations of the reactants and the products as stated by the law of chemical equilibrium at a stage other than that of chemical equilibrium then that ratio is called reaction quotient or concentration quotient.