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At 800 deg C the reaction 2 NO(g) + H2(g) -> N2(g) + H2O(g) was studied. The measured value of (1/2) d[NO]/dt (in M sec⁻¹) at various concentrations was: (i) [NO] = 1.5*10⁻³, [H2] = 4*10⁻³, rate = 4.50*10⁻⁹; (ii) [NO] = 1.5*10⁻³, [H2] = 2*10⁻³, rate = 2.25*10⁻⁹; (iii) [NO] = 3.0*10⁻³, [H2] = 2*10⁻³, rate = 9.00*10⁻⁹. What is the rate of reaction (in M sec⁻¹) when [H2] = 1.5 M and [NO] = 1.0 M?

1. 0.75
2. 0.50
3. 1.50
4. 0.375

Correct answer: 0.75

Solution

Halving [H2] halves the rate (first order in H2); doubling [NO] quadruples the rate (second order in NO), giving rate = k[NO]²[H2] with k = 0.5; at [NO]=1.0, [H2]=1.5 this gives 0.75 M sec⁻¹.

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