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 8.48×10-2 M, the concentration of PH3 will be 2.45×10-2 M after
s have passed.

Answer 1

The decreasing in concentration occurs after 69.0s

Step-by-step explanation:

A first order reaction follows the equation:

Ln [A] = -kt + ln[A]₀

Where [A] is the concentration of the reactant after time t

k is rate constant of the reaction

And [A]₀ is the initial concentration of the reactant

Replacing:

Ln 2.45x10⁻²M = -1.80x10⁻²s⁻¹*t + ln8.48x10⁻²

-3.709 = -1.80x10⁻²s⁻¹*t -2.467

-1.2416 = -1.80x10⁻²s⁻¹*t

69.0s = t

The decreasing in concentration occurs after 69.0s