Problem 1

Nitric oxide reacts with hydrogen at 820 °C:

\ce{2NO(g) + H2(g) -> N2O(g) + H2O(g)}

. Initial rates of

\ce{N2O}

formation were measured at various initial partial pressures (all pressures in torr, times in seconds; do not use concentrations): Exp. 1:

p_{\ce{NO}} = 120.0

p_{\ce{H2}} = 60.0

→ rate

= 8.66 \times 10^{-2}

torr s

^{-1}

; Exp. 2:

p_{\ce{NO}} = 60.0

p_{\ce{H2}} = 60.0

→

2.17 \times 10^{-2}

; Exp. 3:

p_{\ce{NO}} = 60.0

p_{\ce{H2}} = 180.0

→

6.62 \times 10^{-2}

torr s

^{-1}

. (a) Find the rate law and the rate constant. (b) Find the initial rate of disappearance of NO when

p_{\ce{NO}} = 200

torr and

p_{\ce{H2}} = 100

torr. (c) Find the time to halve

p_{\ce{H2}}

when

p_{\ce{NO}} = 800

torr and

p_{\ce{H2}} = 1.0

torr. (d) The proposed mechanism is

\ce{2NO <=> N2O2}

(rate constants

k_1, k_{-1}

) followed by

\ce{N2O2 + H2 ->[k_2] N2O + H2O}

. Derive the rate law using the steady-state approximation for

\ce{N2O2}

, state the condition under which it reduces to the experimental law, and express

k

in terms of

k_1

k_{-1}

k_2

Step 3 of 4: Pseudo-first-order half-life

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k' = k\,p_{\ce{NO}}^{2} = 1.0 \times 10^{-7} \times 800^{2} = 0.064\ \text{s}^{-1};\quad t_{1/2} = \dfrac{\ln 2}{k'} = 10.8\ \text{s}

Analysis

With $p_{\ce{NO}} = 800 \gg p_{\ce{H2}}$ , $p_{\ce{NO}}$ stays essentially constant and the rate law becomes $-dp_{\ce{H2}}/dt = k' p_{\ce{H2}}$ with $k' = k\,p_{\ce{NO}}^2 = 0.064$ s $^{-1}$ , giving $t_{1/2} = \ln 2/k' = 10.8$ s.