Consider proposed mechanism for reaction between nitrogen monoxide and hydrogen gas.Step (i):2NO→N2O2,Step 2:N2O2+H2→N2O+H2O,Step 3:N2O+H2→N2+H2O.If step 2 rate determining step,then rate law for reaction and all the intermediate of reaction will be
Answers
Answer:-
Step 1 ~ NO + NO => N2O2
Rate1 = k1[NO]2
Step 2 ~ N2O2 + H2 => N2O + H2O
Rate2 = k2[N2O2][H2]
Step 3 ~ N2O + H2 => N2 + H2O
Rate3 = k3[N2O][H2]
Overall ~ 2 NO (g) + 2 H2 (g) => N2 (g) + 2 H2O (g)
Rate = k[NO]2[H2]
The rate laws for the three elementary steps will be determined from its chemical equation, or the rate = k [Reactant]coefficient. the rate law is given. The slow (rate determining) step can have the identical rate law because the overall equation.
As seen on top of, the speed law for Step one doesn't up to the rate law, in order that isn't the slow step.
Looking at the speed law for Step two, we have a tendency to see that it contains AN intermediate (N2O2). so as to effectively compare the speed law of Step two with the rate law, we've to urge eliminate [N2O2].
Since we all know that Step one could be a quick step, we will write the equation as a reversible reaction:
NO + NO N2O2
From that, we will write its equilibrium constant:
K = [product]/[reactant] = [N2O2]/[NO]^2
Substituting that into the speed law for
Rate2 = k2[N2O2][H2]
Rate2 = k2 (K[NO]2) [H2]
Combining the constants Dapsang and K as k, we get:
Rate2 = k[NO]^2[H2]
That is the identical because the overall rate law, therefore Step two is that the slow step.
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@GauravSaxena01