[tex] [\text{C}_4 \text{H}_6] \approx 5.2 \times 10^{-3} \; \text{mol} \cdot \text{L}^{-1} [/tex]
Explanation:
From the rate law:
[tex] \text{d} [\text{C}_4 \text{H}_6] /\text{d}t = - 0.014 \cdot [\text{C}_4 \text{H}_6]^{2} [/tex]
where
Note the negative sign in front of the right hand side of the equation. Rearrange the rate law expression to separate the two variables,
[tex] -[\text{C}_4 \text{H}_6]^{-2} \cdot \text{d} [\text{C}_4 \text{H}_6] = 0.014 \cdot \text{d}t [/tex]
Implicitly integrate both sides of the expression (with the power rule) to obtain the general expression (i.e., the one with the arbitary constant "[tex] C [/tex]") for concentration [tex] [\text{C}_4 \text{H}_6] [/tex] at given time [tex] t [/tex]:
[tex] \int(- [\text{C}_4 \text{H}_6]^{-2} )\cdot \text{d} [\text{C}_4 \text{H}_6] = \int 0.014 \cdot \text{d}t [/tex]
[tex] [\text{C}_4 \text{H}_6]^{-1} = 0.014 \cdot \text{t} + C [/tex]
[tex] [\text{C}_4 \text{H}_6] = 1/ (0.014 \cdot \text{t} + C) [/tex]
Now, solve for the value of [tex] C [/tex] for this particular configuration (i.e., with an initial, [tex] t = 0 [/tex] concentration [tex] [\text{C}_4 \text{H}_6] = 0.025 \text{mol}\cdot \text{L}^{-1} [/tex], as seen in the question)
[tex] 1/ (0.014 \cdot \text{t} + C) = 1\; /\; C = [\text{C}_4 \text{H}_6]_{\text{initial}} = 0.025 [/tex]
[tex] C = 1/0.025 = 40.0 [/tex]
Thus, the [tex] \text{C}_4\text{H}_6 [/tex] concentration at time [tex] t [/tex] given an initial [tex] \text{C}_4\text{H}_6 [/tex] concentration of [tex] 0.025 \; \text{mol}\cdot \text{L}^{-1} [/tex] would be
[tex] [\text{C}_4 \text{H}_6](t) = 1/(0.014 \cdot t +40) [/tex]
where again, [tex] t [/tex] the time into the reaction in seconds.
The question is asking for the concentration at [tex] t = 3 \; \text{hrs} = 3 \times 3600 \; \text{seconds} [/tex]
[tex] [\text{C}_4\text{H}_6 ](10800) = 5.2 \times 10^{-3} \; \text{mol} \cdot \text{L}^{-1} [/tex]
Reference:
Lannah Lua, Ciara Murphy, and Victoria Blanchard, "Second-Order Reactions," Libretext Chemistry
