Problem 3 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The value of \(\mathrm{Kc}\) is 64 at \(800 \mathrm{~K}\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) .\) The value of \(K_{\mathrm{c}}\) for the following reaction is : \(\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g})\) (a) \(1 / 64\) (b) 8 (c) \(1 / 4\) (d) \(1 / 8\)

Step-by-Step Solution

Answer

The value of \(K_c\) for the given reaction is \(1/8\) (option d).

Step 1: Understand the Reaction

The given equilibrium constant \(K_c\) is for the reaction \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\), which is the synthesis of ammonia. The \(K_c\) value is 64 at 800 K.

Step 2: Recognize the Reverse Reaction

We are asked to find the \(K_c\) for the reverse of this reaction: \(\text{NH}_3(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g)\). Here, the reaction is written in reverse and involves different stoichiometric coefficients.

Step 3: Apply Equilibrium Constant Rules

Recall that if the equation is reversed, the new equilibrium constant \(K'_c\) is the reciprocal of the original equilibrium constant \(K_c\). Hence, for the reversed reaction, \(K'_c = \frac{1}{K_c} = \frac{1}{64}\).

Step 4: Adjust for Stoichiometry

The modified reaction is not only reversed but also divided by 2. The reaction stoichiometry affects the equilibrium constant in that dividing the coefficients by 2 means you take the square root (or raise to the power of 1/2) of \(K'_c\). Thus, \(K''_c = (K'_c)^{1/2} = \left(\frac{1}{64}\right)^{1/2}.\)

Step 5: Calculate the Final Equilibrium Constant

Now, calculate \(\left(\frac{1}{64}\right)^{1/2} = \frac{1}{8}\). This gives the new equilibrium constant \(K''_c\).

Key Concepts

Reversible Reactions

Reversible reactions play a crucial role in chemical processes by allowing reactions to go in both forward and backward directions. In the provided reaction, \[ \mathrm{N}_{2}( ext{g}) + 3 \mathrm{H}_{2}( ext{g}) \rightleftharpoons 2 \mathrm{NH}_{3}( ext{g}), \] the process is reversible, meaning that ammonia can be formed from nitrogen and hydrogen, and it can also decompose back into these elements. This reversibility is represented by the double-headed arrow (\( \rightleftharpoons \)).
For reversible reactions, equilibrium is achieved when the rate of the forward reaction equals the rate of the backward reaction. This doesn't necessarily mean the concentrations of reactants and products are equal, but they remain steady over time.

• This principles allow for adjustments in concentrations and external conditions to shift the balance of the reaction, which is fundamental to chemical synthesis and industrial processes.
• Understand that reversible reactions are dynamic: they involve continuous movement of elements and compounds, yet might appear static once equilibrium is reached.

Stoichiometry

Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in a chemical reaction. It provides the coefficients that show the proportion in which molecules react and products are formed in a balanced equation.
In the provided reaction, \[ \mathrm{N}_{2}( ext{g}) + 3 \mathrm{H}_{2}( ext{g}) \rightleftharpoons 2 \mathrm{NH}_{3}( ext{g}), \] stoichiometry tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This information is critical when calculating reaction yields and the equilibrium constant.

• When reversing or adjusting a reaction, like dividing the stoichiometric coefficients by 2, the equilibrium constant needs to be adjusted accordingly.
• This adjustment often involves taking a mathematical operation such as a square root, depending on how the stoichiometry is modified.

Stoichiometry ensures that you understand not just how reactants transform during reactions, but also how these transformations can be quantitatively evaluated and predicted, making it a central pillar of chemistry.

Chemical Equilibrium

Chemical equilibrium is a state where the concentrations of all reactants and products remain constant over time because the forward and reverse reactions occur at the same rate. It's an essential concept when considering reversible reactions.
In the case of the ammonia synthesis reaction, a specific equilibrium constant \( K_c \) represents the ratio of product concentrations to reactant concentrations at equilibrium.

• The value of \( K_c = 64 \) at 800 K means that at this temperature, the concentration of ammonia is favored when nitrogen and hydrogen are mixed.
• However, reversing the reaction or changing stoichiometry will alter \( K_c \) since the positions of reactants and products switch.

Understanding equilibrium is crucial because it provides insight into how different factors like temperature, pressure, and concentration affect the position of equilibrium. This comprehension allows chemists to manipulate conditions to favor the synthesis or breakdown of desired chemicals, optimizing industrial processes and even biological systems.