JEE Chemistry 2023 Question with Solution

Given: $N_2 + 3H_2 \rightleftharpoons 2NH_3, \quad K_1 = 4 \times 10^5$ $N_2 + O_2 \rightleftharpoons 2NO, \quad K_2 = 1.6 \times 10^{12}$ $H_2 + \frac{1}{2}O_2 \rightleftharpoons H_2O, \quad K_3 = 1.0 \times 10^{13}$

Find: The equilibrium constant for $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$

From the given working,

$$K_{\text{eq}} = \frac{K_2 \times K_3^3}{K_1}$$

Substituting the values,

$$K_{\text{eq}} = \frac{(1.6 \times 10^{12}) \times (1.0 \times 10^{13})^3}{4 \times 10^5}$$ $$K_{\text{eq}} = \frac{1.6 \times 10^{12} \times 10^{39}}{4 \times 10^5} = 4 \times 10^{33}$$

Therefore, the required nearest integer in the form _____ \times 10^{-33} is $4$.

Given: The three equilibrium reactions with constants $K_1$, $K_2$ and $K_3$.

Find: The constant for $2NH_3 + \frac{5}{2}O_2 \rightleftharpoons 2NO + 3H_2O$.

Use the equilibrium-constant rules:

1. Reversing a reaction inverts its constant.
2. Multiplying a reaction by a number raises the constant to that power.
3. Adding reactions multiplies their constants.

The solution uses:

• reverse of the ammonia formation reaction, giving a factor $\frac{1}{K_1}$
• one copy of the nitric oxide formation reaction, giving $K_2$
• three copies of the water formation reaction, giving $K_3^3$

Hence,

$$K_{\text{eq}} = \frac{K_2 K_3^3}{K_1}$$

Now evaluate it:

$$K_{\text{eq}} = \frac{(1.6 \times 10^{12})(1.0 \times 10^{13})^3}{4 \times 10^5}$$ $$= \frac{1.6 \times 10^{12} \times 10^{39}}{4 \times 10^5}$$ $$= 4 \times 10^{33}$$

So the asked integer coefficient is $4$.