Answer:
The concentration of cyclobutane after 875 seconds is approximately 0.000961 M
Explanation:
The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M
The final concentration of cyclobutane, [[tex]A_t[/tex]] = 0.00119 M
The time for the reaction, t = 455 seconds
Therefore, the Rate Law for the first order reaction is presented as follows;
[tex]\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }[/tex]
Therefore, we get;
[tex]k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}} {-t }[/tex]
Which gives;
[tex]k = \dfrac{\text{ ln} \dfrac {0.00119}{0.00150}} {-455} \approx 5.088 \times 10^{-4}[/tex]
k ≈ 5.088 × 10⁻⁴ s⁻¹
The concentration after 875 seconds is given as follows;
[[tex]A_t[/tex]] = [A₀]·[tex]e^{-k \cdot t}[/tex]
Therefore;
[[tex]A_t[/tex]] = 0.00150 × [tex]e^{5.088 \times 10^{-4} \times 875}[/tex] = 0.000961
The concentration of cyclobutane after 875 seconds, [[tex]A_t[/tex]] ≈ 0.000961 M