MCAT General Chemistry Question 68: Answer and Explanation

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Question: 68

8. The following data shown in the table were collected for the combustion of the theoretical compound XH4:

XH4 + 2 O2 → XO2 + 2 H2O

Trial[XH4]initial (M)[O2]initial (M)
10.60.612.4
20.62.449.9
31.22.4198.3

What is the rate law for the reaction described here?

  • A. rate = k[XH4][O2]
  • B. rate = k[XH4][O2]2
  • C. rate = k[XH4]2[O2]
  • D. rate = k[XH4]2[O2]2

Correct Answer: C

Explanation:

In the first two trials, the concentration of XH4 is held constant, while the concentration of O2 is multiplied by 4. Because the rate of the reaction is also increased by a factor of approximately 4, oxygen must be a first-order reactant. In the last two trials, the concentration of O2 is held constant while the concentration of XH4 is doubled. Because the rate of the reaction is increased by a factor of approximately 4 again, XH4 must be a second-order reactant. Based on this, we can conclude that the experimental rate law is k[XH4]2[O2].

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