Reaction of nitrogen monoxide and hydrogen A student collects the data below for the reaction: 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) Trial [NO]i [H2]i Initial Rate (M/s) 1 0.050 M 0.050 M 0.0917 2 0.10 M 0.050 M 0.367 3 0.10 M 0.10 M 0.734 Determine the rate law for the reaction. multiple choice 1.Rate = k[NO][H2] 2.Rate = k[NO]4[H2]2 3.Rate = k[NO]2[H2]2 4.Rate = k[NO]2[H2] 5.The rate law cannot be determined from the given information.
Reaction of nitrogen monoxide and hydrogen A student collects the data below for the reaction: 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) Trial [NO]i [H2]i Initial Rate (M/s) 1 0.050 M 0.050 M 0.0917 2 0.10 M 0.050 M 0.367 3 0.10 M 0.10 M 0.734 Determine the rate law for the reaction. multiple choice 1.Rate = k[NO][H2] 2.Rate = k[NO]4[H2]2 3.Rate = k[NO]2[H2]2 4.Rate = k[NO]2[H2] 5.The rate law cannot be determined from the given information.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Reaction of nitrogen monoxide and hydrogen
A student collects the data below for the reaction:
2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g)
| Trial | [NO]i | [H2]i | Initial Rate (M/s) |
| 1 | 0.050 M | 0.050 M | 0.0917 |
| 2 | 0.10 M | 0.050 M | 0.367 |
| 3 | 0.10 M | 0.10 M | 0.734 |
Determine the rate law for the reaction.
multiple choice
1.Rate = k[NO][H2]
2.Rate = k[NO]4[H2]2
3.Rate = k[NO]2[H2]2
4.Rate = k[NO]2[H2]
5.The rate law cannot be determined from the given information.
Expert Solution
Step 1
If we look at the trial 1 and 2, the concentration of A is doubled and B remains same. So, on doubling the concentration of A, the rate becomes 4 times of the initial rate. That is 0.0917 becomes 4×0.0917 = 0.367.
Hence the order of reaction with respect to A is 2.
Step 2
If we look at the trial 2 and 3, the concentration of B is doubled and A remains same. So, on doubling the concentration of B, the rate becomes 2 times of the initial rate. That is 0.367 becomes 2×0.367 = 0.734.
Hence the order of the reacton with respect to B is 1.
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