Question

Determine the average rate of change of B from t = 0s to t =392s.

A⟶2B

Time(s) ---- Concentration of A (M)
0 0.720
196 0.430
392 0.140

Answer 1

1) Find the average rate of disappearance of A

$Rate 1 (A) = -\frac{\Delta C}{\Delta t}= -\frac{0.430-0.720}{196-0}=\frac{0.29}{196}=1.48\times 10^{-3} \frac{mol}{L\times s}$

$Rate2(A)= -\frac{\Delta c}{\Delta t}=\frac{0.140-0.430}{392-196}= \frac{0.29}{196}=1.48\times10^{-3}\frac{mol}{L\times s}$

Average rate of A = $1.48\times10^{-3}\frac{mol}{L\times s}$

Since A-> 2B, then to find the average rate of appearance of B we need to multiply the average rate of disappearance of A by 2:

Rate of B =$1.48\times 10^{-3}\times 2= 2.96\times 10^{-3}\frac{mol}{L\times s}$