Q.For the reaction : N2

Concept: Equilibrium Constant Calculation

The equilibrium constant for a reaction changes predictably when the stoichiometric coefficients are altered or the reaction is reversed.

1. If a reaction is multiplied by a factor $n$, the new equilibrium constant is the original constant raised to the power $n$.
2. If a reaction is reversed, the new equilibrium constant is the reciprocal of the original constant.

Given the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ has an equilibrium constant $K_c$.

• (i) $2N_2(g) + 6H_2(g) \rightleftharpoons 4NH_3(g)$: This reaction is the original reaction multiplied by 2. Thus, its equilibrium constant is $K_c^2$. …

The equilibrium constant, $K_c$, for a reversible reaction at a given temperature is a fixed value that describes the ratio of products to reactants at equilibrium. Its value depends on the specific stoichiometry of the balanced chemical equation. When the chemical equation for a reaction is modified, the expression for its equilibrium constant also changes in a specific, predictable way. Understanding these rules is crucial for relating the equilibrium constants of different forms of the same reaction.

Let's consider the given reaction and its equilibrium constant:

Original reaction: $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$

Equilibrium constant: $K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$

Here are the rules for how $K_c$ changes when a reaction is modified:

• Reversing a reaction: If a reaction is reversed, the new equilibrium constant is the reciprocal of the original equilibrium constant. If $\text{A} \rightleftharpoons \text{B}$ has $K_c$, then $\text{B} \rightleftharpoons \text{A}$ has $1/K_c$.
• Multiplying coefficients by a factor: If the stoichiometric coefficients of a reaction are multiplied by a factor $n$, the new equilibrium constant is the original equilibrium constant raised to the power $n$. If $\text{A} \rightleftharpoons \text{B}$ has $K_c$, then $n\text{A} \rightleftharpoons n\text{B}$ has $K_c^n$.
• Dividing coefficients by a factor: If the stoichiometric coefficients of a reaction are divided by a factor $n$ (which is equivalent to multiplying by $1/n$), the new equilibrium constant is the original equilibrium constant raised to the power $1/n$. If $\text{A} \rightleftharpoons \text{B}$ has $K_c$, then $\frac{1}{n}\text{A} \rightleftharpoons \frac{1}{n}\text{B}$ has $K_c^{1/n}$.

Now, let's apply these rules to each reaction in Column I.

1. Reaction (i): $2\text{N}_2\text{(g)} + 6\text{H}_2\text{(g)} \rightleftharpoons 4\text{NH}_3\text{(g)}$

   • Compare this reaction to the original reaction: Original: $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ Reaction (i): $2\text{N}_2\text{(g)} + 6\text{H}_2\text{(g)} \rightleftharpoons 4\text{NH}_3\text{(g)}$
   • We can see that all stoichiometric coefficients in the original reaction have been multiplied by a factor of 2.
   • According to the rule for multiplying coefficients, the new equilibrium constant will be $K_c$ raised to the power of 2.
   • New equilibrium constant for (i) = $K_c^2$.
   • This matches option (d) in Column II.

2. Reaction (ii): $2\text{NH}_3\text{(g)} \rightleftharpoons \text{N}_2\text{(g)} + 3\text{H}_2\text{(g)}$

   • Compare this reaction to the original reaction: Original: $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ Reaction (ii): $2\text{NH}_3\text{(g)} \rightleftharpoons \text{N}_2\text{(g)} + 3\text{H}_2\text{(g)}$
   • This reaction is the reverse of the original reaction. The products of the original reaction are now the reactants, and the reactants are now the products. …