Question

The decomposition of phosphine

\[\ce{4PH3_{(g)} -> P4_{(g)} + 6H2_{(g)}}\]

has the rate law, Rate = k [PH₃].

The rate constant is 6.0 × 10⁻⁴ s⁻¹ at 300 K and activation energy is 3.05 × 10⁵ J mol⁻¹. What is the value of rate constant at 310 K? (R = 8.314 J K⁻¹ mol⁻¹)

Answer 1

Given:

Reaction: \[\ce{4PH3_{(g)} -> P4_{(g)} + 6H2_{(g)}}\]

Rate law: Rate = k [PH₃] → First-order reaction

k1 = 6.0 × 10⁻⁴ s⁻¹ at T₁ = 300 K

T₂ = 310 K

E_a = 3.05 × 10⁵ J mol⁻¹

R = 8.314 J mol⁻¹ K⁻¹

Use the two-point Arrhenius equation:

`ln(k_2/k_1) = E_a/R ((T_2 - T_1)/(T_1T_2))`

`ln(k_2/(6.0 xx 10^-4)) = (3.05 xx 10^5)/8.314 ((310 - 300)/(300 xx 310))`

`ln(k_2/(6.0 xx 10^-4)) = 36675.5 ((10)/(93000))`

`ln(k_2/(6.0 xx 10^-4)) = 36675.5 xx 1.0753 xx 10^-4`

`ln(k_2/(6.0 xx 10^-4)) = 3944`

`(k_2/(6.0 xx 10^-4)) = e^3.944`

= 51.6

⇒ k₂ ​= 6.0 × 10⁻⁴ × 51.6

≈ 0.03096 s⁻¹

∴ k₂ = 3.10 × 10⁻⁴ s⁻¹ at 310 K