For the reaction 2C → A + B, 0.01 mol of A is formed during the first 15 seconds of the reaction. Assuming that the rate of reaction remains constant for two minutes, which of the following statements is true?

Question

Quizplus · Accepted Answer

According to the stoichiometry of the reaction, for every 2 moles of C consumed 1 mole of A and 1 mole of B are produced.

From the given information, we know that:

0.01 mol of A is formed in the first 15 seconds

Thus, during this time,

2C → 0.01 mol A + 0.01 mol B

After simplifying the above equation,

C → 0.005 mol A + 0.005 mol B

As the reaction rate remains constant for 2 minutes,

0.01 mol of A would be formed in the next 15 seconds,

0.01 mol of A would be formed in the next 15 seconds after that,

0.01 mol of A would be formed in the next 15 seconds after that, and so on, until 2 minutes have passed.

Therefore,

0.01 mol A/15 sec x 8 x 2 = 0.08 mol A would be formed after 2 minutes

From the stoichiometry of the reaction,

0.08 mol A is accompanied by 0.08 mol B

Thus,

After 2 minutes, 0.08 moles of B were produced (option A is correct)

As 0.01 mol of A is formed every 15 seconds,

0.01 mol of A is formed in 2 x 60/15 = 8 intervals of 15 seconds each

Thus,

C must have reacted with A and B to form 8 x 2 = 16 moles of A

From the stoichiometry of the reaction,

16 moles of A are accompanied by 16/2 = 8 moles of C

Thus,

After 2 minutes, 0.16 moles of C were consumed (option B is correct)

Therefore, the correct answer is (C) both a and b.

For the Reaction 2C → a + B, 0