The reaction of Nitric Oxide (NO(g)) with molecular hydrogen (H2(g)) results in

Question

The reaction of Nitric Oxide (NO(g)) with molecular hydrogen (H2(g)) results in the production of molecular Nitrogen and water as follows; 2 NO(g) + 2 H2(g) --> N20(g) + 2 H2O(g) The experimentally determined rate-law expression for this reaction is first order in H2(g) and second order in NO(g). a.) Is the reaction as written consistent with the experimentally determined rate-law? b.) An alternative mechanism for the reaction is; k1 2 NO(g) <=> N2O2(g) [fast] k-1 k2 H2(g) + N2O2(g) --> N2O(g) + H2O(g) k3 H2(g) + N2O(g) --> N2(g) + H2O(g) Show that this mechanism is consistent with the experimental rate law.

Explanation / Answer

yes its consistent with the experimentally determined rate-law in case of NO. mechanism is as follows. The intermediates play the key role in finding the overall rate law from mechanism. First consider the reaction steps as elementary reactions and set up set rate laws. Elementary means, that order of reaction is equal to the stoichiometric coefficient of reactant: 2 NO ?? N2O2 r1= k1f · [NO]² - k1r · [N2O2] N2O2 + H2 ? N2O + H2O r2 = k2·[N2O2]·[H2] N2O + H2 ? N2 + H2O r3= k3·[N2O]·[H2] Then set up the material balances for the intermediates d[N2O2] /dt = r1 - r2 = k1f·[NO]² - k1r·[N2O2] - k2·[N2O2]·[H2] d[N2O]/dt = r2 - r3 = k2·[N2O2]·[H2] - k3·[N2O]·[H2] Assume that the intermediate concentration is small due to their high reactivity. Hence their rate of change is negligible small. (Bodenstein theorem). k1f·[NO]² - k1r·[N2O2] - k2·[N2O2]·[H2] = 0 k2·[N2O2]·[H2] - k3·[N2O]·[H2] = 0 Solve for the intermediate concentrations: [N2O2] = k1f·[NO]² / ( k1r + k2·[H2] ) [N2O] = (k2/K3)·[N2O2] = (k1f·k2/K3)·[NO]² / ( k1r + k2·[H2] ) Next formulate the rate of change for one of the products (it doesn't matter which one, result is the same) and insert intermediate concentrations calculated above d[N2]/dt = r3 = k3·[N2O]·[H2] = k1f·k2·[NO]²·[H2] / ( k1r + k2·[H2] ) = k1f· [NO]²·[H2] / ( (k1r/k2) + [H2] ) Last line represents the rate law of the over all reaction: 2NO + 2H2 ? N2 + 2H2O From reaction equation you see that d[NO]/dt = -2d[N2]/dt d[H2]/dt = -2d[N2]/dt d[H2O]/dt = 2d[N2]/dt From rate law you see that rate of overall reaction is 2nd order with respect to nitric oxide. The reaction order with respect to hydrogen depends on its concentration. For small concentrations [H2] > (k1r/k2) rate of reaction is independent from [H2].

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