Problem 97 - A Complete Resource Book in Chemistry for JEE Main

Question

If the equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at 750 is 49 , then the equilibrium constant for the reaction, \(\mathrm{NH}_{3}(\mathrm{~g})=\) \(1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})\) at the same temperature will be (a) \(1 / 49\) (b) 49 (c) \(1 / 7\) (d) \(49^{2}\)

Step-by-Step Solution

Answer

The equilibrium constant for the new reaction is \( \frac{1}{7} \).

Step 1: Understand the Reactions

We are given two reactions. The forward reaction is \( \text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g) \) with an equilibrium constant \( K = 49 \). The reverse reaction provided is \( 2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g) \), and its equilibrium constant will be the reciprocal, i.e., \( K' = \frac{1}{49} \), because the reaction is reversed.

Step 2: Apply Reaction Coefficients

The problem asks for the equilibrium constant when the reaction equation \( \text{NH}_3(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \) is considered. According to the rules of equilibrium constants, if we multiply the stoichiometric coefficients in a reaction by a number, we raise \( K \) to the power corresponding to the reciprocal of that number. Here, the reaction is divided by 2, so the new \( K'' = (\frac{1}{49})^{1/2} = \frac{1}{7} \).

Key Concepts

reverse reaction

When a chemical reaction is presented, it can proceed in both directions: the forward reaction and the reverse reaction. In the forward reaction, reactants turn into products. Conversely, in the reverse reaction, products convert back into reactants. In the example provided, the forward reaction is the synthesis of ammonia: \( \text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g) \). This reaction can be reversed, which is the decomposition of ammonia: \( 2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g) \). The equilibrium constant, \( K \), for the forward reaction was given as 49. The equilibrium constant for the reverse reaction, \( K' \), is the reciprocal, \( \frac{1}{49} \). This reciprocal relationship happens because reversing a reaction flips the ratio of products to reactants.

reaction coefficients

Reaction coefficients are the numerical values placed before the chemical formulas in a balanced chemical equation. They indicate the proportion of molecules or moles involved in the reaction. For example, in the synthesis of ammonia represented by \( \text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g) \), the coefficients are 1 for \( \text{N}_2 \), 3 for \( \text{H}_2 \), and 2 for \( \text{NH}_3 \). These coefficients reflect the stoichiometric ratios necessary for the reaction to occur as described.

• If we change these coefficients, it can significantly affect the calculation of the equilibrium constant.
• Doubling or halving coefficients can change the power to which the equilibrium constant is raised.

In equilibrium constant calculations, we use these coefficients to ascertain how altering them affects the stability and growth of reactants and products.

stoichiometric coefficients

Stoichiometric coefficients play a crucial role in chemical equations and equilibrium calculations. They represent the exact amounts of substances that react and form in a balanced reaction. These coefficients ensure that the law of conservation of mass is upheld, as they help balance the number of atoms for each element on both sides of a reaction.
For instance, in the decomposition of ammonia given by \( 2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g) \), stoichiometric coefficients dictate how much \( \text{NH}_3 \), \( \text{N}_2 \), and \( \text{H}_2 \) are involved.

• When modifying stoichiometric coefficients, such as dividing them by a particular number, we need to adjust equilibrium constant calculations accordingly.
• This precise alteration is shown by raising the equilibrium constant to a power.

Therefore, correctly interpreting stoichiometric coefficients is fundamental when calculating the equilibrium constant for a reaction that has been altered.

equilibrium constant calculation

The equilibrium constant, often denoted as \( K \), provides insight into the ratio of the concentration of products to reactants at equilibrium for a given chemical reaction. Calculating \( K \) is essential for predicting the directionality and extent of a reaction.
In the exercise provided, when the original forward reaction was given as \( K = 49 \), acquiring the reverse reaction constant required taking the reciprocal, \( K' = \frac{1}{49} \).
When dealing with adjusted stoichiometric coefficients, as shown in the modified reaction \( \text{NH}_3(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \), further calculation steps are needed. The original \( K' \) is raised to the power corresponding to the inverse of the ratio applied to the coefficients.

• This example involved taking \( K' = \frac{1}{49} \) and raising it to \( \frac{1}{2} \), resulting in \( K'' = \left( \frac{1}{49} \right)^{\frac{1}{2}} = \frac{1}{7} \).

Such adjustments are crucial for accurately determining \( K \) in reactions with altered coefficients.