Question #d31d8

1 Answer
Feb 5, 2016

Answer:

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"#.