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notka56 [123]
3 years ago
7

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
Chemistry
2 answers:
Ksenya-84 [330]3 years ago
4 0

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

maks197457 [2]3 years ago
3 0
<span>Answer: 5.32 x 10⁻³ M


Explanation:


1) The rate law for a first order reaction is:
</span><span />

<span>r = - d [A] / dt = k[A]
</span><span />

<span>2) When you integrate you get:
</span><span />

<span> [ A] = Ao x e ^(-kt)
</span><span />
Remember that here A is PH₃

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

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