Question

In: Chemistry

The activation energy for a particular reaction is 84 kJ/mol. By what factor will the rate...

The activation energy for a particular reaction is 84 kJ/mol. By what factor will the rate constant increase when the temperature is increased from 50 oC to 72 oC?

Expert Solution

According to Arrhenius Equation , K = A e ^{-Ea /
RT}

Where

K = rate constant

T = temperature

R = gas constant = 8.314 J/mol-K

E_a = activation energy

A = Frequency factor (constant)

Rate constant, K = A e ^{- Ea / RT}

log K = log A - ( E_a / 2.303RT ) ---(1)

If we take rate constants at two different temperatures, then

log K = log A - ( E_a / 2.303RT ) --- (2)

& log K' = log A - (E_a / 2.303RT’) ---- (3)

Eq (3 ) - Eq ( 2 ) gives

log ( K' / K ) = ( E_a / 2.303 R ) x [ ( 1/ T ) - ( 1 / T' ) ]

Given T = 50oC = 50+273 = 323 K

T' = 72 oC = 72+273 = 345 K

Ea = 84 kJ /mol = 84000 J/mol

Plug the values we get

log ( K' / K ) = (84000 / (2.303x8.314) ) x [ ( 1/ 323 ) - ( 1 /345 ) ]

= 0.866

K'/K = 10^0.866

= 7.35

K' = 7.35 x K

Therefore the required factor is 7.35

queen_honey_blossom answered 2 years ago

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