Answer:
Rate law: = [tex]k[CS_2]^1[/tex]
Explanation:
Given:
t [tex][CS_2][/tex]
0.100 [tex]2.7 \times 10^{-7}[/tex]
0.080 [tex]2.2 \times 10^{-7}[/tex]
0.055 [tex]1.5\times 10^{-7}[/tex]
0.044 [tex]1.2\times 10^{-7}[/tex]
Rate law for the given reaction: [tex]k[CS_2]^n[/tex]
Where, n is the order of the reaction.
Divide rate 1 with rate 3
[tex]\frac{0.100}{0.055} =\frac{k[CS_2 (1)]^n}{k[CS_2 (3)]^n} \\\frac{0.100}{0.055} =\frac{k[2.7 \times 10^{-7}]^n}{k[1.5\times 10^{-7}]^n}\\1.81=[1.8]^n\\ n=1[/tex]
So, rate law = [tex]k[CS_2]^1[/tex]
Answer:
[tex]r=-3.73x10^{5}s^{-1} [CS_2][/tex]
Explanation:
Hello,
In this case, a linealization helps to choose the rate law for the decomposition of CS₂ as it is generalized via:
[tex]r=-k[CS_2]^{n}[/tex]
Whereas [tex]n[/tex] accounts for the order of reaction, which could be computed by linealizing the given data using the following procedure:
[tex]-ln(r)=ln(k[CS_2]^{n})\\-ln(r)=ln(k)+ln([CS_2]^{n})\\-ln(r)=ln(k)+n*ln([CS_2])[/tex]
Therefore, on the attached picture you will find the graph and the lineal equation wherein the slope is the order of the reaction and the y-axis intercept the natural logarithm of the rate constant. In such a way, the order of reaction is 1 and the rate constant is:
[tex]ln(k)=12.83\\k=exp(12.83)\\k=3.73x10^{5}s^{-1}[/tex]
Best regards.
