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 4 of 4: Steady-state derivation

← Previous Next →

p_{\ce{N2O2}} = \dfrac{k_1 p_{\ce{NO}}^2}{k_{-1} + k_2 p_{\ce{H2}}}\ \Rightarrow\ \mathrm{Rate} = \dfrac{k_1 k_2\,p_{\ce{NO}}^2 p_{\ce{H2}}}{k_{-1} + k_2 p_{\ce{H2}}}\ \xrightarrow{k_{-1} \gg k_2 p_{\ce{H2}}}\ \dfrac{k_1 k_2}{k_{-1}}\,p_{\ce{NO}}^2 p_{\ce{H2}};\quad k = \dfrac{k_1 k_2}{k_{-1}}

Analysis

Steady state on $\ce{N2O2}$ : $k_1 p_{\ce{NO}}^2 = (k_{-1} + k_2 p_{\ce{H2}})p_{\ce{N2O2}}$ . The rate $\tfrac{1}{2}dp_{\ce{N2O}}/dt$ form reduces to the observed second-order-in-NO law when $k_{-1} \gg k_2 p_{\ce{H2}}$ (fast pre-equilibrium), with $k = k_1 k_2/k_{-1}$ .