Question

At $527{}^o\text{C}$, the reaction given below has $K_c=4$
$N{H}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)$

What is the value of $K_p$ for the reaction given below?
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

• A. $16\times (800R{)}^2$
• B. ${\left(\frac{800R}{4}\right)}^{-2}$
• C. ${\left(\frac{1}{4\times 800R}\right)}^2$
• D. None of these

Answer 1

The correct option is C ${\left(\frac{1}{4\times 800R}\right)}^2$

Given: $527{}^o\text{C}=800\text{K}$ and
$N{H}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right);K_c=4$

Reversing the reaction and multipling by 2, we get
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right);K_c=(\frac{1}{4}{)}^2$
Here, $△n_g=\text{mole of product (g) - moles of reactant (g)}=-2$

Since,
$K_p=K_c(RT{)}^{△n_g}$

$\Rightarrow K_p={\left(\frac{1}{4}\right)}^2(800R{)}^{-2}$

$\Rightarrow K_p={\left(\frac{1}{4\times 800R}\right)}^2$