Answer to Question #141159 in General Chemistry for mike

Question

The rate expression for the reaction:
5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l)
is Rate = k[Br-] [BrO3-] [H+]2

Assuming that all variables except concentration are kept constant, predict the effect on the rate of reaction if:
a). [Br-] doubles, (other reactant concentrations unchanged)

b). [BrO3-] halves, (other reactant concentrations unchanged)

c). [H+] triples, (other reactant concentrations unchanged)

Answer 1

a). Rate1 = k[Br^-][BrO₃^-][H⁺]²

Rate2 = k[2Br^-][BrO₃^-][H⁺]²

$\frac{Rate1}{Rate2} = \frac{k[Br-][BrO3-][H+]^2}{k[2Br-][BrO3-][H+]^2}$

$\frac{Rate1}{Rate2} = \frac{k[Br-][BrO3-][H+]^2}{2k[Br-][BrO3-][H+]^2} = \frac{1}{2}$

The rate will doubles.

b). Rate1 = k[Br^-][BrO₃^-][H⁺]²

Rate2 = k[Br^-][1/2BrO₃^-][H⁺]²

$\frac{Rate1}{Rate2} = \frac{k[Br-][BrO3-][H+]^2}{k[Br-][\frac{1}{2}BrO3-][H+]^2}$

$\frac{Rate1}{Rate2} = \frac{k[Br-][BrO3-][H+]^2}{\frac{1}{2}k[Br-][BrO3-][H+]^2} = 2$

The rate will halves.

c). Rate1 = k[Br^-][BrO₃^-][H⁺]²

Rate2 = k[Br^-][BrO₃^-][3H⁺]²

$\frac{Rate1}{Rate2} = \frac{k[Br-][BrO3-][H+]^2}{k[Br-][BrO3-][3H+]^2}$

$\frac{Rate1}{Rate2} = \frac{k[Br-][BrO3-][H+]^2}{9k[Br-][BrO3-][H+]^2} = \frac{1}{9}$

The rate will become 9 times more.