Consider the following reaction: 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g) The rate law

Question

Consider the following reaction:

2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)


The rate law for this reaction is:

Rate = k [NO(g)]2 [H2(g)]


At a certain temperature when [NO(g)] = 0.0823 M and [H2(g)] = 0.0977 M, the rate of this reaction is 5.78e+01 M s-1. What is the reaction rate when the concentration of H2(g) is doubled, to 0.1954 M, while the concentration of NO(g) is 0.0823 M?



a) The reaction rate is 8.73e+04 M s-1.
b) The reaction rate is 5.80e+01 M s-1.
c) The reaction rate is 2.89e+01 M s-1.
d) The reaction rate is 1.16e+02 M s-1.
e) The reaction rate is 4.37e+04 M s-1.

Please explain why!!!! thank you !

Explanation / Answer

b) The reaction rate is 1.16e+02 M s-1. Rate = k [NO(g)]2 [H2(g)] As the concentration oh H2 double rate becomes 4 times

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