Question

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) 45 g of the first substance was present; 1 hr later, only 10 g of it remained.
Find an expression that gives the amount of the first substance present Q(t) at any time t.

Q(t) = ?

What is the amount present after 2 hr? (Round your answer to two decimal places.)
_________g

Answer 1

Answer:

$Q (t) = \frac{90}{7t +2 }$

$Q(2) = 5.63 \ g$

Step-by-step explanation:

Let Q(t) be the first substance at time t. We know that the rate of change is directly proportional to $Q^2$.

Hence we can write it as:

$\frac{dQ}{dt} \propto Q^2\\\\\frac{dQ}{dt} = kQ^2$

Step 1: Integrating the function to form an equation:

$\frac{dQ}{dt} = kQ^2\\\\\frac{dQ}{Q^2} = k \ dt\\\\\int \frac{dQ}{Q^2} = \int k \ dt\\\\- \frac{1}{Q} = kt + C$   ----------------------------------------------- equation (1)

Step 2: Substitute the value Q(0) = 45 in equation 1 to find value of C.

$-\frac{1}{Q} = kt + C\\\\-\frac{1}{45} = k(0) + C\\\\-\frac{1}{45} = C\\\\\\-\frac{1}{Q} = kt - \frac{1}{45}$---------------------------------------------- equation (2)

Step 3: Substitute the value Q(1) = 10 in equation 2 to find the value of k.

$-\frac{1}{Q} = kt - \frac{1}{45}\\\\-\frac{1}{10} = k(1) - \frac{1}{45}\\\\-\frac{1}{10} - \frac{1}{45} = k \\\\\frac{2-9}{90} = k\\\\-\frac{7}{90} = k$

Step 4: Form the equation and simplify it.

$-\frac{1}{Q} = -\frac{7}{90} t - \frac{1}{45}\\ \\\frac{1}{Q} = \frac{7}{90}t + \frac{2}{90}\\\\\frac{1}{Q} = \frac{7t + 2}{90}\\\\\\\bullet \ cross - multiply \ the \ equation \\\\Q = \frac{90}{7t + 2}$

Step 5: To find the amount of substance after 2 hours, we substitute the value of t with 2.

$Q = \frac{90}{7t + 2}\\\\Q = \frac{90}{7(2) + 2}\\\\Q = \frac{90}{14+ 2}\\\\Q = \frac{90}{16}\\\\Q = 5.625 \approx 5.63$

The amount of substance after 2 hours is 5.63 grams.