Respuesta :
Explanation:
As the given reaction is as follows.
[tex]A + B \rightarrow 2C + D[/tex]
at t = 0 [tex]C_{Ao}[/tex] [tex]3C_{Ao}[/tex] 0 0
at t = t' [tex]C_{Ao} - C_{Ao}X_{Ao}[/tex] [tex]3C_{Ao} - C_{Ao}X_{A}[/tex] [tex]2X_{A}C_{Ao}[/tex] [tex]C_{Ao}X_{A}[/tex]
[tex]-r_{A}[/tex] = [tex]\frac{dC_{A}}{dt}[/tex] = [tex]kC_{A}C_{B}[/tex]
= [tex]kC_{Ao}(1 - X_{A})C_{Ao}(3 - X_{A})[/tex]
= [tex]kC^{2}_{Ao}(1 - X_{A})(3 - X_{A})[/tex]
[tex]\frac{dC_{Ao}(1 - X_{A})}{dt}[/tex] = [tex]kC^{2}_{Ao}(1 - X_{A})(3 - X_{A})[/tex]
[tex]-C_{Ao}\frac{dX_{A}}{dt}[/tex] = kC^{2}_{Ao} (1 - X_{A})(3 - X_{A})[/tex]
[tex]\int_{0}^{0.5}\frac{dX_{A}}{(1 - X_{A})(3 - X_{A})}[/tex] = [tex]kC_{Ao}\int_{0}^{10 min}dt[/tex]
[tex]\int_{0}^{0.5}[\frac{1}{2(1 - X_{A})} - \frac{1}{2(3 - X_{A})}] dX_{A}[/tex] = [tex]kC_{Ao} \times (10 - 0)[/tex]
[tex]\frac{1}{2}[-ln (1 - X_{A})]^{0.5}_{0} + \frac{1}{2}[ln(3 - X_{A})]^{0.5}_{0}[/tex] = [tex]kC_{Ao} \times 10[/tex]
[tex]\frac{1}{2}[ln 1 - ln 0.5]^{0.5}_{0} + \frac{1}{2} \times (\frac{2.5}{3})[/tex] = [tex]kC_{Ao} \times 10[/tex]
0.2554 = [tex]kC_{Ao} \times 10[/tex]
Given, [tex]C_{Ao}[/tex] = 0.01 mol/L
k = [tex]\frac{0.2554 L mol^{-1} min^{-1}}{0.01 \times 10}[/tex]
= 2.554 [tex]L mol^{-1} min^{-1}[/tex]
Thus, we can conclude that value of given reaction rate constant is 2.554 [tex]L mol^{-1} min^{-1}[/tex].
