Experiment [O 2 (g)]( M ) [NO(g)]( M ) Rate ( M /s) 1 0.0130 0.0130 1.66e-02 2 0

Question

Experiment [O2(g)](M) [NO(g)](M) Rate (M/s) 1 0.0130 0.0130 1.66e-02 2 0.0130 0.0260 6.63e-02 3 0.0260 0.0130 3.31e-02 4 0.0260 0.0260 1.33e-01 Experiment [O2(g)](M) [NO(g)](M) Rate (M/s) 1 0.0130 0.0130 1.66e-02 2 0.0130 0.0260 6.63e-02 3 0.0260 0.0130 3.31e-02 4 0.0260 0.0260 1.33e-01 The reaction, O2(g) +2 NO(g) right arrow 2 NO2(g), was studied at acertain temperature with the following results: If the rate law for this reaction is, Rate =k [O2(g)] [NO(g)]2, what is the valueof the rate constant? Please show the equation used when finding rate constant. thank you.

Explanation / Answer

With the rate law, you're almost done. Let's use the first experiment to find k. Plug in the valuesthat you know (namely, the rate and concentrations): 1.66e-2 = k (0.013) (0.013)2 Then solve for k k =7555 What are the units? Well, if the units for rate are mol/Lsec and when multiplied by(mol/L)3 (from the concentrations), k must be in mol/Lsec = k mol/L mol/L mol/L k = L2/mol2sec So k = 7555 L2/mol2sec

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