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Solve The energy of activation for a first-order reaction is 104 kJ/mol. The rate constant at 25°C is 3.7 × 10–5 s –1. What is the rate constant at 30°C? (R = 8.314 J/K mol) - Chemistry

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The energy of activation for a first-order reaction is 104 kJ/mol. The rate constant at 25°C is 3.7 × 10–5 s –1. What is the rate constant at 30°C? (R = 8.314 J/K mol)

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Solution

Given:

Activation energy (Ea) = 104 kJ mol-1 = 104 × 103 J mol-1

Rate constant (k1) = `3.7 xx 10^-5 "s"^-1`

Temperatures; T1 = 25 + 273 = 298 K, T= 30 + 273 = 303 K

R = 8.314 J K-1 mol-1

To find:

Rate constant (k2) at 30°C

Formula:

`"log"_10 "k"_2/"k"_1 = "E"_"a"/(2.303"R") (("T"_2 - "T"_1)/("T"_2"T"_1))`

Calculation: 

`"log"_10 "k"_2/(3.7 xx 10^-5 "s"^-1) = (104 xx 10^3 "J"  "mol"^-1)/(2.303 xx 8.314 "J"  "K"^-1 "mol"^-1) ((303 "K" - 298 "K")/(303 "K" xx 298 "K"))`

∴ `"log"_10  "k"_2/(3.7 xx 10^-5 "s"^-1) = 104000/(2.303 xx 8.314) xx 5/(303 xx 298)`

∴ `"log"_10 "k"_2/(3.7 xx 10^-5 "s"^-1) = 0.301`

∴ `"k"_2/(3.7 xx 10^-5 "s"^-1)` = antilog(0.301) = 2.00

k2 = `2.00 xx 3.7 xx 10^-5 "s"^-1 = 7.4 xx 10^-5 "s"^-1`

The rate constant of the reaction is `7.4 xx 10^-5 "s"^-1`.

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APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 6 Chemical Kinetics
Exercise | Q 4.3 | Page 137
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