Question

The balanced equation for the reaction of bromate ion with bromide in acidic solution is $BrO^- + 5Br^- + 6H^+ \rightarrow 3Br_2 + 3H_2O$At a particular instant in time, the rate of disappearance of Br– is 2.0 x 10⁻³ mol/L • s. What is the rate of appearance of Br₂ at the same instant?

Answer 1

Answer:

The rate of appearance of Br₂ at the same instant is 0.0012 mol/L.

Explanation:

$BrO^-+5Br^-+6H^+\rightarrow 3Br_2+3H_2O$

The rate of disappearance of $Br^$:

$-\frac{dBr^-}{dt}=2.0\times 10^{-3} mol/L$

Rate of the reaction in terms of bromide ions : R

$R=-\frac{1}{5}\frac{dBr^-}{dt}$

$R=\frac{1}{5}\times 2.0\times 10^{-3} mol/L=0.0004 mol/L$

Rate of the reaction in terms of bromine gas : R

$R=\frac{1}{3}\frac{dBr_2}{dt}$

$0.0004 mol/L=\frac{1}{3}\frac{dBr_2}{dt}$

$\frac{dBr_2}{dt}$=3\times 0.0004 mol/L=0.0012 mol/L[/tex]

The rate of appearance of Br₂ at the same instant is 0.0012 mol/L.