In a chemical reaction, the rate at which the amount of a reactant changes with time is proportional to the amount present, such that $\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }$, when $\mathrm { t }$ is measured in hours. If there are $63 \mathrm {~g}$ of reactant present when $t$ $= 0$, how many grams will be left after 2 hours? Give your answer to the nearest tenth of a gram.
A) 15.5 g
B) 23.3 g
C) 0.2 g
D) 7.8 g

Question

Quizplus · Accepted Answer

The differential equation $\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }$ is a first-order linear differential equation. The solution to this equation is $y(t) = Ce^{-0.7t}$, where $C$ is a constant. Using the initial condition $y(0) = 63$, we get $C = 63$. Therefore, the amount of reactant present at time $t$ is $y(t) = 63e^{-0.7t}$. After 2 hours, the amount of reactant left is $y(2) = 63e^{-0.7(2)} = 15.5$.

In a Chemical Reaction, the Rate at Which the Amount dydt=−0.7y\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }dtdy​=−0.7y

In a Chemical Reaction, the Rate at Which the Amount $\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }$