Homework 15.65

Arrhenius Equation:

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Homework 15.65

Postby arif_latif_2G » Sun Mar 10, 2019 6:45 pm

Can someone help me through this question:
For the reversible, one-step reaction 2 A <=>B B + C, the forward rate constant for the formation of B is 265 Lmol^-1min^-1 and the rate constant for the reverse reaction is 392 Lmol^-1min^-1. The activation energy for the forward reaction is 39.7 kJmol^-1 and that of the reverse reaction is 25.4 kJmol^-1.
(a) What is the equilibrium constant for the reaction? (b) Is the reaction exothermic or endothermic?
(c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant?

Tam To 1B
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Joined: Fri Sep 28, 2018 12:25 am

Re: Homework 15.65

Postby Tam To 1B » Mon Mar 11, 2019 4:02 pm

(a) In order to find the equilibrium constant, you divide the forward rate constant by the reverse rate constant. K = k1/k'1 = 265/392 = 0.676.
(b) The reaction is endothermic because the activation energy of the forward reaction is higher than the activation energy of the reverse reaction. In an endothermic reaction, the potential energy of the products is higher than the reactants, so a higher activation energy is needed to reach the state of the products.
(c) Raising the temperature increases the rate constant of the forward reaction. In raising k1, the equilibrium constant K also increases.

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