The decomposition of phosphine,PH₃,follows first-order kinetics. 4 PH₃(g) → P₄(g) + 6 H₂(g)
The half-life for the reaction at 550 °C is 81.3 seconds.What percentage of phosphine remains after 195 seconds?

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the given half-life to calculate the rate constant for the reaction. 
The half-life (t1/2) of a first-order reaction is given by the equation:
t1/2 = (0.693/k)
where k is the rate constant.

Substituting the given values, we have:
81.3 s = (0.693/k)
k = 0.008514 s^-1

Now we can use the rate constant to calculate the concentration of phosphine remaining after 195 seconds using the first-order rate equation:
ln([PH3]t/[PH3]0) = -kt
where [PH3]t is the concentration of phosphine at time t, [PH3]0 is the initial concentration, k is the rate constant, and t is time.

Rearranging this equation, we have:
[PH3]t = [PH3]0 * e^(-kt)

Substituting the given values, we have:
[PH3]195 = [PH3]0 * e^(-0.008514*195)
[PH3]195 = [PH3]0 * 0.1878

Taking the ratio of [PH3]195 and [PH3]0 and multiplying by 100 gives us the percentage of phosphine remaining after 195 seconds:
([PH3]195/[PH3]0) * 100 = 0.1878 * 100 = 18.78%

Therefore, the answer is choice C, 19%.

The Decomposition of Phosphine,PH3,follows First-Order Kinetics

The Decomposition of Phosphine,PH₃,follows First-Order Kinetics