Question

At a certain temperature, 0.3411 0.3411 mol of N 2 N2 and 1.661 1.661 mol of H 2 H2 are placed in a 2.50 2.50 L container. N 2 ( g ) 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) N2(g) 3H2(g)↽−−⇀2NH3(g) At equilibrium, 0.2001 0.2001 mol of N 2 N2 is present. Calculate the equilibrium constant, K c Kc .

Answer 1

Answer: The equilibrium constant for the above reaction is 1.31

Explanation:

We are given:

Initial moles of nitrogen gas = 0.3411 moles

Initial moles of hydrogen gas = 1.661 moles

Equilibrium moles of nitrogen gas = 0.2001 moles

For the given chemical reaction:

                   $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial:         0.3411      1.661

At eqllm:     0.3411-x  1.661-3x      2x

Evaluating for 'x', we get:

$\Rightarrow (0.3411-x)=0.2001\\\\\Rightarrow x=0.3411-0.2001=0.141$

Volume of the container = 2.50 L

The expression of $K_c$ for the above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

We are given:

$[NH_3]=\frac{2\times 0.141}{2.50}=0.1128M$

$[N_2]=\frac{0.2001}{2.5}=0.08004M$

$[H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M$

Putting values in above expression, we get:

$K_c=\frac{(0.1128)^2}{0.08004\times (0.4952)^3}\\\\K_c=1.31$

Hence, the equilibrium constant for the above reaction is 1.31