The temperature at which the rate constants of the given below two gaseous reactions become equal is ____ K. (Nearest integer). X arrow Y k₁ = 10⁶e(-30000)/(T)$$X \rightarrow Y \quad k_{1} = 10^{6}e^{\frac{-30000}{T}}$$ P arrow Q k₂ = 10⁴e(-24000)/(T)$$P \rightarrow Q \quad k_{2} = 10^{4}e^{\frac{-24000}{T}}$$ Given: ln 10 = 2.303$\ln 10 = 2.303$

Related Formula

Rate = k[A]$Rate = k[A]$

Core Logic

For a first order reaction, rate is directly proportional to concentration.

Step 1: Setting up equations

Step 2: Solving for k

Dividing equation (1) by (2):

Take natural log on both sides:

Step 3: Calculating half life

Convert to minutes by dividing by 60:

Substitute log values (since ln x = 2.303 x$\ln x = 2.303 \log x$, the 2.303 cancels out):

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Related Formula

Core Logic

Since r = k[A]$r = k[A]$, the reaction follows first-order kinetics. The half-life (50%$50\%$ decomposition) is t1/2 = 120 minutes$t_{1/2} = 120\text{ minutes}$.

Step 1: Calculating for 90% decomposition

For 90%$90\%$ completion of the reaction, [A]₀ = 100$[A]_0 = 100$ and [A]ₜ = 100 - 90 = 10$[A]_t = 100 - 90 = 10$.

Rounding off to the nearest integer, we get 399 minutes$399\text{ minutes}$.

Pattern Recognition

For a first order reaction, t90% ≈ 3.32 × t50%$t_{90\%} \approx 3.32 \times t_{50\%}$. 120 × 3.32 = 398.4$120 \times 3.32 = 398.4$, so roughly 399$399$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Core Logic

Consider a general gas phase reaction: A arrow B + C$A \rightarrow B + C$

Initial (t=0$t=0$): Pᵢ 0 0$\quad P_i \quad\quad 0 \quad\quad 0$ At time t$t$: Pᵢ - x x x$\quad P_i - x \quad\quad x \quad\quad x$

Total pressure at time t$t$:

x = Pₜ - Pᵢ$x = P_t - P_i$

Partial pressure of A at time t$t$ (PA$P_A$): PA = Pᵢ - x$P_A = P_i - x$

For a first-order reaction:

Here, P₀ = Pᵢ$P_0 = P_i$ and the pressure of the reactant at time t$t$ is PA$P_A$.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics