Answer to Question #108955 in General Chemistry for Gabby

Question

When Pt metal is used as a catalyst for the previous reaction, we see that the mechanism changes and the reaction is much faster. The activation energy is found to be 98.4 kJ mol-1 with the catalyst at room temperature. How much would you have to raise the temperature to get the catalyzed reaction to run 100 times faster than it does at room temperature with the catalyst? Please answer in °C.

Answer 1

$ln{\frac {k_2} {k_1}}= {\frac {E_a} {R}}({\frac {1} {T_1}} - {\frac {1} {T2}})$

E_a = 98.4 kJ*mol^-1 = 98*10³ J*mol^-1

R = 8.314 J*mol^-1*K^-1

T₁ = room temperature = 20 ⁰C = 293 K

${\frac {k_2} {K_1}}=100$

T₂ = x

$ln100= {\frac {98400} {8.314}}({\frac {1} {293}} - {\frac {1} {x}})$

x = $({\frac {1} {293}}-{\frac {8.314*ln100} {98400}})^{-1}=330.7≈331$ K

331 - 273 = 58 ⁰C