The rate constant of a reaction is 6.4 × 10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter your answer in scientific notation.

k₂ = 5 x 10⁻² s⁻¹

Explanation:

Using the Clausis - Clapyron Equation ...

ln(k₁/k₂) = ΔH/R(T₁-T₂/T₁·T₂) => lnk₂ = lnk₁ + ΔH/R(T₁-T₂/T₁·T₂)

k₁ = 6.4 x 10⁻³s⁻¹

ΔH = 33.6 Kj/mole

R = 0.008314 Kj/mol·K

T₁ = 25°C = 298K

T₂ = 75°C = 348K

lnk₂ = ln(6.4 x 10⁻³s⁻¹) + (33.6kj/mol·K/0.008314Kj/mol·K)·[(348K - 298K)/(348K)(298K)] = -3.10s⁻¹

k₂ at 75°C = exp(-3.10) = 5 x 10⁻²s⁻¹ (faster k-value at higher temp)

the answer is true.

i think it is true since you can test the answer and can use observations to complete it.