Question

important: if you use this answer in later parts, use the full unrounded value in your calculations. calculate the reaction rate using the expression determined in part a: rate = -\frac{1}{2}\frac{\deltahbr}{\delta t}= -\frac{1}{2}\frac{hbr_2 - hbr_1}{t_2 - t_1}= -\frac{1}{2}\frac{0.471 m - 0.520 m}{22.0 s - 0.0 s}= 1.1\times10^{-3} m/s part c if the volume of the reaction vessel in the previous part was 1.25 l, what amount of br₂ (in moles) was formed during the first 15.0 s of the reaction? express your answer in moles to two significant figures. view available hint(s) n = mol

Explanation:

Step1: Find the reaction - rate of HBr

First, we need to find the rate of change of HBr concentration. Assume the rate of the reaction is still related to the rate of change of HBr as in the previous part. Let's assume the rate of the reaction is $r =-\frac{1}{2}\frac{\Delta[HBr]}{\Delta t}$. We need to find the rate of the reaction in the first 15.0 s. But we don't have the [HBr] values at 0 s and 15 s directly. However, if we assume the rate is constant (or we use the average rate concept), we can use the rate from the previous calculation (if the reaction is of a simple order and conditions are similar). Let's assume the rate $r = 1.1\times10^{-3}M/s$.

Step2: Determine the rate of formation of $Br_2$

For the reaction involving HBr and $Br_2$, the stoichiometric relationship is used. If the reaction is $2HBr
ightarrow H_2 + Br_2$, the rate of formation of $Br_2$ is $r_{Br_2}=\frac{1}{2}r_{HBr}$. Since $r =-\frac{1}{2}\frac{\Delta[HBr]}{\Delta t}$, the rate of formation of $Br_2$ is $r_{Br_2}= 1.1\times10^{-3}M/s$ (because of the stoichiometry of the reaction $2HBr
ightarrow H_2+Br_2$, the rate of formation of $Br_2$ is half of the rate of disappearance of HBr in magnitude).

Step3: Calculate the change in concentration of $Br_2$

We know that rate $r_{Br_2}=\frac{\Delta[Br_2]}{\Delta t}$. So, $\Delta[Br_2]=r_{Br_2}\times\Delta t$. Substituting $r_{Br_2}=1.1\times 10^{-3}M/s$ and $\Delta t = 15.0s$, we get $\Delta[Br_2]=1.1\times10^{-3}M/s\times15.0s = 1.65\times10^{-2}M$.

Step4: Calculate the moles of $Br_2$

We know that $n = M\times V$. Given $V = 1.25L$ and $M=\Delta[Br_2]=1.65\times 10^{-2}M$, then $n=1.65\times10^{-2}mol/L\times1.25L=2.0625\times10^{-2}mol\approx2.1\times 10^{-2}mol$.

Answer:

$2.1\times 10^{-2}$ mol