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**Solve**

The rate constant of a reaction at 500°C is 1.6 × 10^{3} M^{–1} s ^{–1}. What is the frequency factor of the reaction if its activation energy is 56 kJ/mol?

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#### Solution

**Given: **

Rate constant (k) = `1.6 xx 10^3 "M"^-1 "s"^-1`,

Temperature (T) = 500 + 273 = 773 K,

Activation energy (E_{a}) = 56 kJ mol^{-1 }= 56 × 10^{3 }J mol^{-1}

**To find: **

frequency factor (A)

**Formula:**

K = `"Ae"^((-"E"_"a")/("RT")`

**Calculation: **

Substituting the given values

`1.6 xx 10^3 "M"^-1 "s"^-1 = "A" xx "e"^(((-56 xx 10^3 "J" "mol"^-1)/(8.314 "J" "K"^-1 "mol"^-1xx 773 "K")))`

∴ `(1.6 xx 10^3 "M"^-1 "s"^-1)/"A" = "e"^(((-56000)/(8.314 xx 773)))`

∴ `"log" ((1.6 xx 10^3 "M"^-1 "s"^-1)/"A") = (-56000)/(8.314 xx 773 xx 2.303)`

∴ `(1.6 xx 10^3 "M"^-1 "s"^-1)/"A"` = antilog(-3.7836)

∴ `(1.6 xx 10^3 "M"^-1 "s"^-1)/"A" = 1.646 xx 10^-4`

∴ A = `((1.6 xx 10^3 "M"^-1 "s"^-1)/(1.646 xx 10^-4)) = 9.72 xx 10^6 "M"^-1 "s"^-1`

The frequency factor of reaction is `9.72 × 10^-6 "M"^-1 "s"^-1`.

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