Question

In the first 22.0 s of this reaction, the concentration of HBr dropped from 0.590 M to 0.465 M . Calculate the average rate of the reaction in this time interval.

Answer 1

This is an incomplete question, here is a complete question.

Consider the following reaction:

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

In the first 22.0 s of this reaction, the concentration of HBr dropped from 0.590 M to 0.465 M . Calculate the average rate of the reaction in this time interval.

Answer : The average rate of reaction is, $6.25\times 10^{-3}M/s$

Explanation :

The given chemical reaction follows:

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

The average rate of the reaction for disappearance of HBr is given as:

$\text{Average rate of disappearance of HBr}=-\frac{\Delta [HBr]}{\Delta t}$

where,

$C_2$ = final concentration of HBr = 0.465 M

$C_1$ = initial concentration of HBr = 0.590 M

$\Delta t$ = change in time = 22.0 s

Putting values in above equation, we get:

$\text{Average rate of reaction}=-\frac{(0.465-0.590)M}{20.0s}\\\\\text{Average rate of reaction}=6.25\times 10^{-3}M/s$

Hence, the average rate of reaction is, $6.25\times 10^{-3}M/s$