Answer:
[tex]r=0.0341\frac{M}{s}[/tex]
Explanation:
Hello!
In this case, considering the given table, we are able to represent the symbolic rate law as shown below:
[tex]r=k[BF_3]^m[NH_3]^n[/tex]
Thus, by using the following steps, we can find both m (BF3 order of reaction) and n (BF3 order of reaction):
- Experiment 1 and 2 for the calculation of n:
[tex]\frac{r_1}{r_2}=\frac{k[BF_3]_1^m[NH_3]_1^n}{k[BF_3]_2^m[NH_3]_2^n}[/tex]
So we plug in to obtain:
[tex]\frac{0.2130}{0.1065}=\frac{k*0.250^m*0.250^n}{k*0.250^m*0.125^n}\\\\2=\frac{0.250^n}{0.125^n}\\\\2=2^n\\\\log(2)=n*log(2)\\\\n=1[/tex]
So the order of reaction with respect to NH3 is 1.
- Experiment 3 and 4 for the calculation of m
[tex]\frac{r_3}{r_4}=\frac{k[BF_3]_3^m[NH_3]_3^n}{k[BF_3]_4^m[NH_3]_4^n}[/tex]
So we plug in to obtain:
[tex]\frac{0.0682}{0.1193}=\frac{k*0.200^m*0.100^n}{k*0.350^m*0.100^n}\\\\0.57=0.57^m\\\\log(0.57)=m*log(0.57)\\\\m=1[/tex]
So the order of reaction with respect to BF3 is also 1.
Now, we can compute the rate constant by solving for it on any of the experiments there, say experiment 1:
[tex]k=\frac{r_1}{[BF_3][NH_3]} =\frac{0.2130M/s}{0.250M*0.250M}\\\\k=3.41M^{-1}s^{-1}[/tex]
Thus, the initial reaction rate for the 0.50M BF3 and 0.020M NH3 is:
[tex]r=3.41M^{-1}s^{-1}*0.50M*0.020M\\\\r=0.0341\frac{M}{s}[/tex]
Best regards!
