## 15.65 c

$K = \frac{k_{forward}}{k_{reverse}}$

sahajgill
Posts: 20
Joined: Fri Sep 29, 2017 7:06 am

### 15.65 c

15.65 For the reversible, one-step reaction 2A ⇌ B + C, the forward rate constant for the formation of B is 265 L.mol^-1.min^-1 and the rate constant for the reverse reaction is 392 L.mol^-1.min^-1. The activation energy for the forward reaction is 39.7 kJ.mol^-1 and that of the reverse reaction is 25.4 kJ.mol^-1. (c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant?

Can someone please explain to me why the statement "raising the temperature will increase the rate constant of the reaction with the higher activation barrier more than it will the rate constant of the reaction with the lower energy barrier" is true?

Jonathan Tangonan 1E
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

### Re: 15.65 c

The rate constant and the equilibrium constant of endothermic reactions are increasing with the raising temperature (as temperature goes up k and K). This is because when temperature increases the equilibrium shifts towards the right.

Sophie 1I
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.65 c

K=k/k' so if k is bigger the equilibrium constant will increase.