Question

The gas phase decomposition of phosphine at 120 °C PH3(g) 1/4 P4(g) + 3/2 H2(g) is first order in PH3 with a rate constant of 1.80×10-2 s-1. If the initial concentration of PH3 is 3.16×10-2 M, the concentration of PH3 will be M after 99 s have passed

Answer 1

Answer:

5.32*10⁻³M

Explanation:

Given:

Rate constant of the First order reaction, k = 1.80*10-2 s-1

Initial concentration of PH3, [A]₀ = 3.16*10-2 M

Reaction time, t = 99 s

Formula:

For a first order reaction:

$[A] = [A]_{0} e^{-kt}$

where [A] and [A]₀ are concentration of reactant at time t and t = 0

k = rate constant

For the given reaction"

$[A] = 3.16*10^{-2} e^{-1.80*10^{-2} *99} = 5.32*10^{-3} M$

Answer 2

Answer: 5.32 x 10⁻³ M


Explanation:


1) The rate law for a first order reaction is:


r = - d [A] / dt = k[A]


2) When you integrate you get:


[ A] = Ao x e ^(-kt)

Remember that here A is PH₃

3) Plug in the data: Ao = 0.0316M, k = 0.0180 /s, and t = 99s


[PH₃] = 0.0316 M x e ^( - 0.0180/s x 99s) = 5.32 x 10⁻³ M