In a certain chemical reaction, a substance is converted into another substance at a rate proportional to the square of the amount of the first substance present at any time t. Initially (t = 0) 60 g of the first substance was present; 1 hr later, only 35 g of it remained. Find an expression that gives the amount of the first substance present Q (t) at any time t.

Q (t) = 60 / ((5/7) t + 1)

Or

Q (t) = 60 / (0.714t + 1)

Step-by-step explanation:

Let Q (t) be the amount of the first substance present at time t. Then:

dQ/dt = - k*Q²

where k is a positive consstant of proportionality.

This is a separable differential equation:

Separating it we have;

-dQ/Q² = k dt

Integrating both sides:

1/Q - 1/Qo = k*t ... 1

where Qo = Q (0) is the amount of the first substance present at time t = 0

1/Q = k*t + 1/Qo

Q (t) = Qo / (Qo*k*t + 1) ... 2

Given that Qo = 60g, so:

Q (t) = (60) / ((60) * k*t + 1)

And also at t = 1 hr, Q (1 hr) = 35, so from equation 1:

1 / (35) - (1/60) = k * (1 hr)

k = 1/84

Therefore, the equation 2 becomes;

Q (t) = Qo / (Qo*k*t + 1) ... 2

Q (t) = 60 / (60*1/84*t + 1)

Q (t) = 60 / ((5/7) t + 1)

Or

Q (t) = 60 / (0.714t + 1)