Question

section b: answer all the questions
question one (chemical kinetics) 10 marks
for the reaction:
2n₂o₅(𝑙) → 4no₂(𝑔) + o₂(𝑔)
the rate is directly proportional to n₂o₅. at 45 °c, 90% of n₂o₅ reacts in 3600 seconds.
find the value of the rate - constant k.
5
the rate law for the decomposition of n₂o₅(𝑙) is: rate = kn₂o₅ where
k = 6.22×10⁻⁴s⁻¹. calculate half - life of n₂o₅ and the number of seconds it will take for an
initial concentration of n₂o₅ of 0.100m to drop to 0.0100m.
5
question two (organic chemistry) 10 marks

Explanation:

Step1: Determine the order of the reaction

Since the rate is directly proportional to $[N_2O_5]$, it is a first - order reaction.

Step2: For the first part (finding k)

For a first - order reaction, the integrated rate law is $\ln\frac{[A]_0}{[A]_t}=kt$. If 90% of $[N_2O_5]$ reacts in 3600 s, then $[A]_t=(1 - 0.9)[A]_0 = 0.1[A]_0$. Substituting into the integrated rate law: $\ln\frac{[A]_0}{0.1[A]_0}=k\times3600$. $\ln10 = k\times3600$, so $k=\frac{\ln10}{3600}\ s^{-1}\approx6.39\times 10^{-4}\ s^{-1}$.

Step3: Calculate the half - life ($t_{1/2}$)

The formula for the half - life of a first - order reaction is $t_{1/2}=\frac{\ln2}{k}$. Given $k = 6.22\times 10^{-4}\ s^{-1}$, then $t_{1/2}=\frac{\ln2}{6.22\times 10^{-4}\ s^{-1}}\approx1119\ s$.

Step4: Calculate the time for concentration change

Using the integrated rate law $\ln\frac{[A]_0}{[A]_t}=kt$, with $[A]_0 = 0.100\ M$ and $[A]_t=0.0100\ M$. $\ln\frac{0.100}{0.0100}=kt$. $\ln10 = kt$. Since $k = 6.22\times 10^{-4}\ s^{-1}$, then $t=\frac{\ln10}{6.22\times 10^{-4}\ s^{-1}}\approx3698\ s$.

Answer:

For the first part, $k\approx6.39\times 10^{-4}\ s^{-1}$. For the second part, $t_{1/2}\approx1119\ s$ and $t\approx3698\ s$.