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A certain first-order reaction has a rate constant of 2.15×10−2 s−1 at 20 ∘C. What is the value of k at 55 ∘C if Ea = 72.0 kJ/mol ?

User Gareth Jones
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1 Answer

15 votes
15 votes

Answer:


k_2=0.504s^(-1)

Step-by-step explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:


ln((k_2)/(k_1) )=-(Ea)/(R)((1)/(T_2) -(1)/(T_1) )

Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:


ln((k_2)/(0.0215s^(-1)) )=-(72000(J)/(mol))/(8.3145(J)/(mol*K))((1)/(55+273) -(1)/(20+273) ) \\\\ln((k_2)/(0.0215s^(-1)) )=3.154\\\\k_2=0.0215s^(-1)exp(3.154)\\\\k_2=0.504s^(-1)

Regards!

User Sventevit
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