Question

Use the graph to calculate the instantaneous rate of formation of HBr at 50 s

Answer 1

Answer: The instantaneous rate of formation of HBr at 50 s is $1.4\times 10^{-2}M/s$

Explanation:

From the graph,

Initial rate of the $Br_2$ = 1.0 M

Time when the concentration of $Br_2$ is 0.5 M (half the concentration ) = 60 sec

For first order reaction:

Calculating rate constant for first order reaction using half life:

$t_{1/2}=\frac{0.693}{k}$ .....(1)

$t_{1/2}$ = half life period = 60 s

k = rate constant = ?

Putting values in equation 1:

$k=\frac{0.693}{60s}\\\\k=0.01155s^{-1}$

For the given chemical reaction:

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

Rate of the reaction = $-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}$

Negative sign represents the disappearance of the reactants

From the above expression:

$k[Br_2]=-\frac{\Delta [Br_2]}{\Delta t}=\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}$

At 50 seconds, $[Br_2]=0.6 M$

Plugging values in above expression, we get:

$\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=0.01155\times 0.6\\\\\frac{\Delta [HBr]}{\Delta t}=2\times 0.01155\times 0.6=0.01386=1.4\times 10^{-2}M/s$

Hence, the instantaneous rate of formation of HBr at 50 s is $1.4\times 10^{-2}M/s$