Answer: The instantaneous rate of formation of HBr at 50 s is [tex]1.4\times 10^{-2}M/s[/tex]
Explanation:
From the graph,
Initial rate of the [tex]Br_2[/tex] = 1.0 M
Time when the concentration of [tex]Br_2[/tex] is 0.5 M (half the concentration ) = 60 sec
For first order reaction:
Calculating rate constant for first order reaction using half life:
[tex]t_{1/2}=\frac{0.693}{k}[/tex] .....(1)
[tex]t_{1/2}[/tex] = half life period = 60 s
k = rate constant = ?
Putting values in equation 1:
[tex]k=\frac{0.693}{60s}\\\\k=0.01155s^{-1}[/tex]
For the given chemical reaction:
[tex]H_2(g)+Br_2(g)\rightarrow 2HBr(g)[/tex]
Rate of the reaction = [tex]-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}[/tex]
Negative sign represents the disappearance of the reactants
From the above expression:
[tex]k[Br_2]=-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}[/tex]
At 50 seconds, [tex][Br_2]=0.6 M[/tex]
Plugging values in above expression, we get:
[tex]\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=0.01155\times 0.6\\\\\frac{\Delta [HBr]}{\Delta t}=2\times 0.01155\times 0.6=0.01386=1.4\times 10^{-2}M/s[/tex]
Hence, the instantaneous rate of formation of HBr at 50 s is [tex]1.4\times 10^{-2}M/s[/tex]
