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If [NO] = 0.60 M and [H2] = 0.35 M, what is the rate of the reaction H2(g) + 2NO(g) ⟶ N2O(g) + H2O(g)?

a. 1.134 × 10^(-2) mol L^(-1) s^(-1)
b. 2.268 × 10^(-2) mol L^(-1) s^(-1)
c. 3.402 × 10^(-3) mol L^(-1) s^(-1)
d. 5.670 × 10^(-3) mol L^(-1) s^(-1)

1 Answer

7 votes

Final answer:

If [NO] = 0.60 M and [H2] = 0.35 M, the rate of the reaction is a. 1.134 × 10^(-2) mol L^(-1) s^(-1)

Step-by-step explanation:

The rate of a chemical reaction is determined by the rate law, which describes how the rate of the reaction depends on the concentrations of the reactants.

In this question, the rate law for the reaction H2(g) + 2NO(g) ⟶ N2O(g) + H2O(g) is rate = k[NO]^2[H2], where k is the rate constant.

To find the rate of the reaction, we need to substitute the given concentrations into the rate law equation.

Given: [NO] = 0.60 M and [H2] = 0.35 M

Substituting the given concentrations into the rate law equation:

rate = k(0.60)^2(0.35) = k(0.126) = 0.126k

Therefore, the rate of the reaction is 0.126 times the rate constant k.

The correct answer is option a. 1.134 × 10^(-2) mol L^(-1) s^(-1).

User Jpoveda
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