Chemistry Labs

Problem 1

Nitric oxide reacts with hydrogen at 820 °C: 2 NO(g)+HX2(g)→NX2O(g)+HX2O(g)\ce{2NO(g) + H2(g) -> N2O(g) + H2O(g)}. Initial rates of NX2O\ce{N2O} formation were measured at various initial partial pressures (all pressures in torr, times in seconds; do not use concentrations): Exp. 1: pNO=120.0p_{\ce{NO}} = 120.0, pHX2=60.0p_{\ce{H2}} = 60.0 → rate =8.66×10−2= 8.66 \times 10^{-2} torr s−1^{-1}; Exp. 2: pNO=60.0p_{\ce{NO}} = 60.0, pHX2=60.0p_{\ce{H2}} = 60.0 → 2.17×10−22.17 \times 10^{-2}; Exp. 3: pNO=60.0p_{\ce{NO}} = 60.0, pHX2=180.0p_{\ce{H2}} = 180.0 → 6.62×10−26.62 \times 10^{-2} torr s−1^{-1}. (a) Find the rate law and the rate constant. (b) Find the initial rate of disappearance of NO when pNO=200p_{\ce{NO}} = 200 torr and pHX2=100p_{\ce{H2}} = 100 torr. (c) Find the time to halve pHX2p_{\ce{H2}} when pNO=800p_{\ce{NO}} = 800 torr and pHX2=1.0p_{\ce{H2}} = 1.0 torr. (d) The proposed mechanism is 2 NO⇌NX2OX2\ce{2NO <=> N2O2} (rate constants k1,k−1k_1, k_{-1}) followed by NX2OX2+HX2→kX2NX2O+HX2O\ce{N2O2 + H2 ->[k_2] N2O + H2O}. Derive the rate law using the steady-state approximation for NX2OX2\ce{N2O2}, state the condition under which it reduces to the experimental law, and express kk in terms of k1k_1, k−1k_{-1}, k2k_2.
Step 2 of 4: NO disappearance rate
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−dpNOdt=2k pNO2pHX2=2×1.0×10−7×2002×100=0.80 torr s−1-\dfrac{dp_{\ce{NO}}}{dt} = 2k\,p_{\ce{NO}}^{2}p_{\ce{H2}} = 2 \times 1.0 \times 10^{-7} \times 200^{2} \times 100 = 0.80\ \text{torr s}^{-1}
Analysis

The stoichiometry 2 NO+HX2→NX2O+HX2O\ce{2NO + H2 -> N2O + H2O} consumes 2 NO per NX2O\ce{N2O} formed, so the NO decay rate is twice the NX2O\ce{N2O} production rate: 2×0.40=0.802 \times 0.40 = 0.80 torr s−1^{-1}.

Common pitfall. The measured rate is for NX2O\ce{N2O} production; multiply by 2 for NO consumption.