Question

2X (g)+Y(g) +3Z (g) → Products. The rate equation of above reaction is given by: Rate = K [X]¹ [Y]⁰ [Z]². Choose the correct statements

• A. If [z] >> [x] and 75% of X undergoes reaction in 20 sec, then 50% of X will react in 10 sec.
• B. Rate of reaction decreases by reducing the concentration of Y to half of the original value
• C. The half-life of Z increases by increasing its concentration if [x] >> [z]
• D. On doubling the concentration of each of X, Y & Z, rate of reaction becomes 8 times

Answer 1

Answer: (A), (D)

• If [z] >> [x] and 75% of X undergoes reaction in 20 sec, then 50% of X will react in 10 sec.
• On doubling the concentration of each of X, Y & Z, rate of reaction becomes 8 times

Answer 2

The given rate equation is: Rate = $K [X]¹ [Y]⁰ [Z]²$

This means the reaction is first order with respect to X, zero order with respect to Y, and second order with respect to Z.

1. Statement 1: "If [z] >> [x] and 75% of X undergoes reaction in 20 sec, then 50% of X will react in 10 sec." If [z] >> [x], the concentration of Z can be considered constant during the reaction with X. The rate law becomes Rate = $K [X] [Z]₀² = K_{eff} [X]$, where $K_{eff} = K [Z]₀²$ is a constant. This is a pseudo-first-order reaction with respect to X. For a first-order reaction, the integrated rate law is $ln([X]₀/[X]) = K_{eff} t$. Time for 75% reaction: $[X] = [X]₀ - 0.75[X]₀ = 0.25[X]₀$. $ln([X]₀/0.25[X]₀) = K_{eff} t_{75\%}$ $ln(4) = K_{eff} \times 20$ $2 ln(2) = K_{eff} \times 20 \implies K_{eff} = \frac{ln(2)}{10}$ Time for 50% reaction (half-life, $t_{1/2}$): $[X] = [X]₀ - 0.50[X]₀ = 0.50[X]₀$. $ln([X]₀/0.50[X]₀) = K_{eff} t_{50\%}$ $ln(2) = K_{eff} t_{50\%}$ Substitute $K_{eff}$: $ln(2) = \frac{ln(2)}{10} t_{50\%} \implies t_{50\%} = 10$ sec. The statement is correct.

2. Statement 2: "Rate of reaction decreases by reducing the concentration of Y to half of the original value" The rate law is Rate = $K [X]¹ [Y]⁰ [Z]² = K [X] [Z]²$. The rate of reaction is independent of the concentration of Y because the reaction order with respect to Y is 0. Reducing the concentration of Y will not change the rate, assuming [X] and [Z] are constant. The statement is incorrect.

3. Statement 3: "The half-life of Z increases by increasing its concentration if [x] >> [z]" If [x] >> [z], the concentration of X can be considered constant during the reaction with Z. The rate law becomes Rate = $K [X]₀ [Z]² = K_{eff} [Z]²$, where $K_{eff} = K [X]₀$ is a constant. This is a pseudo-second-order reaction with respect to Z. For a second-order reaction (Rate = $k' [A]²$), the half-life is given by $t_{1/2} = \frac{1}{k' [A]₀}$. In this case, the half-life of Z is $t_{1/2, Z} = \frac{1}{K_{eff} [Z]₀} = \frac{1}{K [X]₀ [Z]₀}$. The half-life of Z is inversely proportional to its initial concentration $[Z]₀$. Increasing $[Z]₀$ will decrease the half-life of Z. The statement is incorrect.

4. Statement 4: "On doubling the concentration of each of X, Y & Z, rate of reaction becomes 8 times" Let the initial concentrations be $[X]₀$, $[Y]₀$, and $[Z]₀$. Initial Rate = $K [X]₀¹ [Y]₀⁰ [Z]₀²$. New concentrations are $[X]' = 2[X]₀$, $[Y]' = 2[Y]₀$, $[Z]' = 2[Z]₀$. New Rate = $K [X']¹ [Y']⁰ [Z']² = K (2[X]₀)¹ (2[Y]₀)⁰ (2[Z]₀)²$ New Rate = $K (2[X]₀) (1) (4[Z]₀²)$ New Rate = $8 \times K [X]₀ [Z]₀²$ New Rate = 8 $\times$ Initial Rate. The rate of reaction becomes 8 times. The statement is correct.

The correct statements are 1 and 4.

Explanation: The rate law is Rate = $K [X]¹ [Y]⁰ [Z]²$. Statement 1: Under the condition [Z] >> [X], the reaction is pseudo-first order w.r.t. X. For a first-order reaction, $t_{75\%} = 2 \times t_{50\%}$. Given $t_{75\%} = 20$s, $t_{50\%} = 10$s. Correct. Statement 2: The rate is independent of [Y] as the order w.r.t. Y is 0. Incorrect. Statement 3: Under the condition [X] >> [Z], the reaction is pseudo-second order w.r.t. Z. For a second-order reaction, $t_{1/2} \propto 1/[reactant]₀$. Increasing [Z]₀ decreases $t_{1/2}$. Incorrect. Statement 4: Rate $\propto [X]¹ [Z]²$. Doubling [X], [Y], [Z] changes rate by $(2)^1 \times (2)^0 \times (2)^2 = 2 \times 1 \times 4 = 8$ times. Correct.