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:
[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]
Initial: 0.3411 1.661
At eqllm: 0.3411-x 1.661-3x 2x
Evaluating for 'x', we get:
[tex]\Rightarrow (0.3411-x)=0.2001\\\\\Rightarrow x=0.3411-0.2001=0.141[/tex]
Volume of the container = 2.50 L
The expression of [tex]K_c[/tex] for the above equation follows:
[tex]K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}[/tex]
We are given:
[tex][NH_3]=\frac{2\times 0.141}{2.50}=0.1128M[/tex]
[tex][N_2]=\frac{0.2001}{2.5}=0.08004M[/tex]
[tex][H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M[/tex]
Putting values in above expression, we get:
[tex]K_c=\frac{(0.1128)^2}{0.08004\times (0.4952)^3}\\\\K_c=1.31[/tex]
Hence, the equilibrium constant for the above reaction is 1.31
