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Question: The rate constant of the reaction is $\bigcirc$ 10$^{-5}$mmHg$^{-2}$ s$^{-1}$ $\bigcirc$ 10$^{-6}$...

The rate constant of the reaction is

◯\bigcirc 10−5^{-5}mmHg−2^{-2} s−1^{-1}

◯\bigcirc 10−6^{-6}mmHg−2^{-2} s−1^{-1}

◯\bigcirc 10−7^{-7}mmHg−2^{-2} s−1^{-1}

◯\bigcirc 10−8^{-8}mmHg−2^{-2} s−1^{-1}

A

10−5^{-5}mmHg−2^{-2} s−1^{-1}

B

10−6^{-6}mmHg−2^{-2} s−1^{-1}

C

10−7^{-7}mmHg−2^{-2} s−1^{-1}

D

10−8^{-8}mmHg−2^{-2} s−1^{-1}

Answer

10−7^{-7}mmHg−2^{-2} s−1^{-1}

Explanation

Solution

The rate law is given as dp(N2O)dt=k(pNO)2pH2\frac{dp(\text{N}_2\text{O})}{dt} = k(p_{\text{NO}})^2 p_{\text{H}_2}.

From Experiment 1, (pNO)0=600(p_{\text{NO}})_0 = 600 mmHg, (pH2)0=10(p_{\text{H}_2})_0 = 10 mmHg, t1/2=19.2t_{1/2} = 19.2 s at 827 °C. Since (pNO)0≫(pH2)0(p_{\text{NO}})_0 \gg (p_{\text{H}_2})_0, the reaction is pseudo-first order with respect to H2_2, with effective rate constant k′=k(pNO)02k' = k(p_{\text{NO}})_0^2. The half-life is t1/2=ln⁡2k′t_{1/2} = \frac{\ln 2}{k'}. So, 19.2=ln⁡2k(600)219.2 = \frac{\ln 2}{k(600)^2}, which gives k=ln⁡219.2×6002≈1.00×10−7k = \frac{\ln 2}{19.2 \times 600^2} \approx 1.00 \times 10^{-7} mmHg−2^{-2} s−1^{-1}.

From Experiment 3, (pNO)0=10(p_{\text{NO}})_0 = 10 mmHg, (pH2)0=600(p_{\text{H}_2})_0 = 600 mmHg, t1/2=830t_{1/2} = 830 s at 827 °C. Since (pH2)0≫(pNO)0(p_{\text{H}_2})_0 \gg (p_{\text{NO}})_0, the reaction is pseudo-second order with respect to NO, with effective rate constant for consumption of NO as 2k′′=2k(pH2)02k'' = 2k(p_{\text{H}_2})_0. The half-life for a second-order reaction is t1/2=1keff(pNO)0t_{1/2} = \frac{1}{k_{eff} (p_{\text{NO}})_0}. Here, keffk_{eff} is the rate constant for the decrease in concentration of NO in the second-order rate law −dpNOdt=keff(pNO)2-\frac{dp_{\text{NO}}}{dt} = k_{eff} (p_{\text{NO}})^2. From the given rate law, −dpNOdt=2k(pNO)2pH2-\frac{dp_{\text{NO}}}{dt} = 2k(p_{\text{NO}})^2 p_{\text{H}_2}. With pseudo-second order approximation, −dpNOdt=2k(pNO)2(pH2)0-\frac{dp_{\text{NO}}}{dt} = 2k(p_{\text{NO}})^2 (p_{\text{H}_2})_0. So keff=2k(pH2)0k_{eff} = 2k(p_{\text{H}_2})_0. The half-life is t1/2=12k(pH2)0(pNO)0t_{1/2} = \frac{1}{2k(p_{\text{H}_2})_0 (p_{\text{NO}})_0}. So, 830=12k×600×10830 = \frac{1}{2k \times 600 \times 10}, which gives k=1830×2×600×10≈1.00×10−7k = \frac{1}{830 \times 2 \times 600 \times 10} \approx 1.00 \times 10^{-7} mmHg−2^{-2} s−1^{-1}.

Both experiments give a consistent value of kk around 10−710^{-7} mmHg−2^{-2} s−1^{-1} at 827 °C.