Question

For the reaction A + 3B → 2C, how does the rate of disappearance of B compare to the
rate of production of C?​

Answer 1

Explanation:

$\frac{1}{3} \times \: rate \: of \: disappearence \:of \: a \\ b = \frac{1}{2} \times rate \: of \: appearence \: of \: c$

Answer 2

The given reaction is A + 3B → 2C, which indicates that one molecule of A and three molecules of B react to form two molecules of C. To determine how the rate of disappearance of B compares to the rate of production of C, we need to consider the stoichiometry of the reaction and the rate law.

• The stoichiometry of the reaction shows that for every one molecule of A, three molecules of B react to form two molecules of C. This means that the rate of disappearance of B is three times the rate of disappearance of A, and the rate of production of C is two times the rate of disappearance of A.
• The rate law for this reaction can be expressed as follows: Rate = k[A]^x[B]^y, where k is the rate constant, x and y are the reaction orders with respect to A and B, respectively. The overall reaction order is x + y.
• Since the reaction order with respect to B is 1, and the reaction order with respect to A is also 1, the rate of disappearance of B is directly proportional to the concentration of B, and the rate of production of C is directly proportional to the concentration of A.
• Therefore, as the concentration of B decreases, the rate of disappearance of B will decrease, and as the concentration of A decreases, the rate of production of C will decrease.
• In terms of numerical comparison, the rate of disappearance of B is three times faster than the rate of production of C, as three molecules of B are required to produce two molecules of C.
• In summary, the rate of disappearance of B is directly proportional to the concentration of B and is three times faster than the rate of production of C, which is directly proportional to the concentration of A.

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