The following mechanism has been suggested for the reaction between nitrogen mon
ID: 791267 • Letter: T
Question
The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen:NO(g) + NO(g) ? N2O2(g) (fast)N2O2(g) + O2(g) ? 2NO2(g) (slow)According to this mechanism, the experimental rate law is Answer second-order in NO and zero-order in O2. second-order in NO and first-order in O2. first-order in NO and first-order in O2. first-order in NO and second-order in O2. first-order in NO and zero-order in O2. The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen:NO(g) + NO(g) ? N2O2(g) (fast)N2O2(g) + O2(g) ? 2NO2(g) (slow)According to this mechanism, the experimental rate law is The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen:NO(g) + NO(g) ? N2O2(g) (fast)N2O2(g) + O2(g) ? 2NO2(g) (slow)According to this mechanism, the experimental rate law is second-order in NO and zero-order in O2. second-order in NO and first-order in O2. first-order in NO and first-order in O2. first-order in NO and second-order in O2. first-order in NO and zero-order in O2. second-order in NO and zero-order in O2. second-order in NO and first-order in O2. first-order in NO and first-order in O2. first-order in NO and second-order in O2. first-order in NO and zero-order in O2.Explanation / Answer
second-order in NO and first-order in O?.
Explanation:
The slowest step determines the overall rate of reaction. That means overall rate of reaction is the same as rate of the second reaction:
rate = k??[N?O?]?[O?]
The rate of the first reaction, which is a reversible reaction, is much faster. Therefore it will be (almost) in equilibrium state. Hence the concentration of reactants and products are related by the equilibrium equation:
K? = [N?O?]/[NO]
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