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 1 of 4: Orders and rate constant

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\mathrm{Rate} = k\,p_{\ce{NO}}^{a}p_{\ce{H2}}^{b};\quad \dfrac{8.66}{2.17} = 2^{a} \Rightarrow a = 2;\quad \dfrac{6.62}{2.17} = 3^{b} \Rightarrow b = 1;\quad k = \dfrac{8.66 \times 10^{-2}}{120^{2} \times 60} = 1.00 \times 10^{-7}\ \text{torr}^{-2}\,\text{s}^{-1}

Analysis

Doubling $p_{\ce{NO}}$ at fixed $p_{\ce{H2}}$ quadruples the rate (order 2 in NO); tripling $p_{\ce{H2}}$ triples it (order 1 in $\ce{H2}$ ). Substitution into any experiment gives $k = 1.00 \times 10^{-7}$ torr $^{-2}$ s $^{-1}$ .