Extraordinary Differential Equations #10

Calculus Level 3

Consider two reactants A and B in the following reaction in a 1:1 stoichiometric ratio to give product C : $A+B \rightarrow C$ The reaction is first order with respect to both A and B . You are given that the rate of formation of C is thus directly proportional to the concentrations of both A and B , i.e. we have the following differential equation: $Rate=\frac{d[C]}{dt}=k[A][B]$ where $[C]$ is the concentration of C (etc.), and $k$ is a proportionality constant called the rate constant . For this particular reaction $k=2.50\times10^{-4} \text{ dm}^{3} \text{ mol}^{-1} \text{ s}^{-1}$ . If the initial concentrations of A and B are 3 $\text{mol dm}^{-3}$ and 1 $\text{mol dm}^{-3}$ respectively, and that there is no C in the beginning, calculate the concentration of C in $\text{mol dm}^{-3}$ after 1 hour.

(This problem is part of the set Extraordinary Differential Equations .)

0.34 1.5E-3 0.044 0.88 7.5E-4 1.76

1 solution

Wee Xian Bin
Feb 4, 2017

First we rewrite the rate law into: $Rate=\frac{d[C]}{dt}=k(3-[C])(1-[C])$ which can be converted into separable form with partial fractions: $(\frac{1}{[C]-3}-\frac{1}{[C]-1}) \text{ }d[C]=2k \text{ }dt$ $\ln{([C]-3)}-\ln{([C]-1)}=2kt+c$ $\ln({\frac{[C]-3}{[C]-1}})=2kt+c$ where $c$ is the arbitrary constant of integration. To obtain the value of $c$ consider the initial condition $[C]=0$ when $t=0$ , thus $c=\ln3$ . After some algebraic manipulation we get $[C]=\frac{1-e^{(5\times10^{-4})t}}{1/3-e^{(5\times10^{-4})t}}$ Then the required concentration of C after an hour (3600 s) is $=0.88 \text{ mol dm}^{-3}$ .

Extraordinary Differential Equations #10

1 solution

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