Answer :
This is an incomplete question, here is a complete question.
Consider the following reaction:
[tex]2HBr(g)\rightarrow H_2(g)+Br_2(g)[/tex]
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, [tex]6.25\times 10^{-3}M/s[/tex]
Explanation :
The given chemical reaction follows:
[tex]2HBr(g)\rightarrow H_2(g)+Br_2(g)[/tex]
The average rate of the reaction for disappearance of HBr is given as:
[tex]\text{Average rate of disappearance of HBr}=-\frac{\Delta [HBr]}{\Delta t}[/tex]
where,
[tex]C_2[/tex] = final concentration of HBr = 0.465 M
[tex]C_1[/tex] = initial concentration of HBr = 0.590 M
[tex]\Delta t[/tex] = change in time = 22.0 s
Putting values in above equation, we get:
[tex]\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[/tex]
Hence, the average rate of reaction is, [tex]6.25\times 10^{-3}M/s[/tex]