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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.

Respuesta :

Answer:

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)

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