# Question d31d8

Feb 5, 2016

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

#### Explanation:

The relative amounts of $\text{B}$ and $\text{C}$ depend on how fast they are formed.

The rate laws for the formation of $\text{B}$ and $\text{C}$ 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

$\text{B}$ is formed 3.32 times as fast as $\text{C}$, so you will get 3.32 mol of $\text{B}$ for every 1 mol of $\text{C}$,

"% 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"#.