Question

Question #d31d8

Answer 1

The relative amounts are `77 % "B"` and `23 % "C"`.

Explanation:

The relative amounts of `"B"` and `"C"` depend on how fast they are formed.

The rate laws for the formation of `"B"` and `"C"` are:

`"rate"_"B" = k_1["A"]`
`"rate"_"C" = k_2["A"]`

`"rate"_(text(B))/"rate"_C = (k_1color(red)(cancel(color(black)(["A"]))))/(k_2color(red)(cancel(color(black)(["A"])))) = k_1/k_2 = (1.26 × 10^(-4) "s"^(-1))/(3.8 × 10^(-5) "s"^(-1)) = 3.32/1`

`"B"` is formed 3.32 times as fast as `"C"`, so you will get 3.32 mol of `"B"` for every 1 mol of `"C"`,

`"% B" = "moles of B"/"moles of A + moles of B" × 100 % = "3.32 mol"/"3.32 mol + 1 mol" × 100 % = (3.32 color(red)(cancel(color(black)("mol"))))/(4.32 color(red)(cancel(color(black)("mol")))) × 100 % = 77 %`

The mixture contains `77 %"B"` and `23 % "C"`.