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What would happen to the rate of a reaction with law rate k NO2H2 if the concentration of H2 were halved?
The rate would be four times larger. Impossible.
What would happen to the rate of a reaction with rate law rate k NO2H2 if NO were halved?
In the rate law given as rate = k[NO2][H2], the concentration of NO does not appear, so the rate of the reaction is independent of its concentration. Therefore, if the concentration of NO were halved, it would have no effect on the rate of the reaction. The reaction rate would remain unchanged as long as the concentrations of NO2 and H2 remain constant.
What would happen to the rate of a reaction with rate law rate k NO2H2 if the concentration of H2 were halved?
If the concentration of H2 is halved, it would also halve the rate of the reaction, assuming H2 is a reactant in the rate law. This is because the rate law is directly proportional to the concentrations of reactants.
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