Question

14) The rate constant of a first-order reaction is 3.00 × 10^−4 s^−1 at 350.°C. If the activation energy is 149 kJ/mol, calculate the temperature at which its rate constant is 7.60 × 10^−4 s^−1.

_____ *C

Answer 1

Given

k₁ = 3.00 $\times$ 10^-4 s^-1

k₂ = 7.60 $\times$ 10^-4 s^-1

T₁ = 350 ⁰C = 350 + 273.15 = 623.15 K

T₂ = ?

Ea= 149 kJ/mol = 149000 J/mol

R = 8.314 J/mol.K

Now we have the formula

ln(k₂ /k₁) = -(Ea/R).(1/T₂ - 1/T₁ )

by rearranging we get

-Rln(k₂ /k₁)/ Ea = 1/T₂ - 1/T₁

-Rln(k₂ /k₁)/ Ea + 1/T₁ = 1/T₂

T₂ = 1 / {1/T₁ - Rln(k₂ /k₁)/ Ea}

T₂ = EaT₁ / {Ea - T₁R.ln(k₂ /k₁) }

so by using above values and formula we get

T₂ = 149000 J/mol $\times$ 623.15 K / { 149000 J/mol - 623.15 K $\times$ 8.314 J/mol.K $\times$ (ln 7.60 $\times$ 10^-4 s^-1/3.00 $\times$ 10^-4 s^-1 )

= 92849350 J.K/mol / { 149000 J/mol - 3590.86)

= 638.53 K

= 638.53 - 273.15

= 365.38 ⁰C

So our asnwer is 365 ⁰C