Question

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 10 grams of C is formed in 5 minutes. How much is formed in 10 minutes? (Round your answer to one decimal place.)

Answer 1

Answer:  17.6 grams

Step-by-step explanation:

As the problem tells us, the velocity of the reaction is proportional to the product of the quantities of A and B that have not reacted, so from this we get the next equation:

                                                       V = k[A][B]

where [A] represents the remaining amount of A, and [B] represents the remaining amount of B. To solve this equation we have to represent it through a differential equation, which is:

                                              dx/dt = k[α - a(t)][β - b(t)]         (1)

where,

k: velocity constant

a(t): quantity of A consumed in instant t

b(t): quantity of B consumed in instant t

α: initial quantity of A

β: initial quantity of B

Now we need to define the equations for a(t) and b(t), and for this we are going to use the law of conservation of mass by Lavoisier, with which we can say that the quantity of C in a certain instant is equal to the sum of the quantities of A and B that have reacted. Therefore, if we need M grams of A and N grams of B to form a quantity of M+N of C, then we can say that in a certain time, the consumed quantities of A and B are given by the following equations:

                                       a(t) = ( M/M+N) · x(t)

                                       b(t) = (N/M+N) · x(t)

where,

x(t): quantity of C in instant t

So for this problem we have that for 1 gram of B, 2 grams of A are used, therefore the previous equations can be represented as:

                                       a(t) = (2/2+1) · x(t) = 2/3 x(t)

                                       b(t) = (1/2+1) · x(t) = 1/3 x(t)

Now we proceed to resolve the differential equation (1) by substituting values:

                                         dx/dt = k[α - a(t)][β - b(t)]  

                                        dx/dt = k[40 - 2x/3][50 - x/3]

                                         dx/dt = k/9 [120 - 2x][150 - x]

We use the separation of variables method:

                                      dx/[120-2x][150-x] = k/3 · dt

We integrate both sides of the equation:

                                     ∫dx/(120-2x)(150-x) = ∫kdt/9

                                     ∫dx/(15-x)(60-x) = kt/9 + c

Now, to integrate the left side of the equation we need to use the partial fraction decomposition:

                                    ∫[1/90(120-2x) - 1/180(150-x)] = kt/9 + c

                                      1/180 ln(150-x/120-2x) = kt/9 + c

                                           (150-x)/(120-2x) = Ce^{20kt}

Now we resolve by taking into account that x(0) = 0, and x(5) = 10,

for x(0) = 0 ,             (150-0)/(120-0) = Ce^{20k(0)} , C = 1.25

for x(5) = 10 ,           (150-10)/(120-(2·10)) = 1.25e^{20k(5)} , k ≈ 113 · 10^{-5}

Now that we have the values of C and k, we have this equation:

                           (150-x)/(120-2x) = 1.25e^{226·10^{-4}t}

and we have to clear by x, obtaining:

               x(t) = 150 · (1 - e^{226·10^{-4}t} / 1 - 2.5e^{226·10^{-4}t})

Therefore the quantity of C that will be formed in 10 minutes is:

           x(10) = 150 · (1 - e^{226·10^{-4}(10)} / 1 - 2.5e^{226·10^{-4}(10)})

                                            x(10) ≈ 17.6 grams