Question

If for the equilibrium :
$N{H}_{2}COON{H}_4\left(s\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)+CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$
the value of Kp at 800 K is 27 x $2^{\lambda /2}$ and the equilibrium pressure is 22 atm. Value of $\lambda$ is :

• A. 12
• B. 21
• C. 10
• D. 11
• E. 4

Answer 1

Answer: (B)

21

The ratio of the number of moles of $N_2\left(g\right),H_2\left(g\right),CO\left(g\right)and{O}_2\left(g\right)$ is $1:3:1:0.5$. The total number of moles are 4.5. If P represents the partial pressure of nitrogen, then the total pressure will be 4.5 P which is equal to 22 atm. Hence, $P=\frac{22}{4.5}=4.89$ atm. The equilibrium constant expression is $K_p=P_{N_2\left(g\right)}\times P_{H_2\left(g\right)}^3\times P_{CO\left(g\right)}\times P_{O_2\left(g\right)}^{0.5}$. Substitute values in the above expression.
$27\times 2^{\lambda /2}=P_{4.89}\times P_{4.89}^3\times P_{4.89}\times P_{4.89}^{0.5}$
Hence, $\lambda =21$.