Question

For a Reaction that has positive activation energy, would you expect increasing the temperature to increase or decrease the reaction rate? Please explain your answer in using "rate=K[A]^x[B]^y" and "K=Ae*-Ea/rT"

Answer 1

we know that

K = Ae^-(Ea / RT)

ln K = ln A - (Ea / RT)

now

at two different temperatures

ln K1 = ln A - ( Ea / RT1)

ln K2 = ln A - (Ea / RT2)

solving we get

ln ( K2 / K1) = (Ea / R) ( 1/T1 - 1/T2)

given

Ea = +ve

also

the temperature is increased

so

T2 > T1

1/T2 < 1/ T1

1/T1 - 1/T2 > 0

also

Ea > 0

so

ln ( K2 / K1) > 0

so

K2 > K1

so

the rate constant has increased

now

rate = K [A]^x [B]^y

as K increased

the rate also will increase

so

the reaction rate is increased as the temperature is increased