Sapling Question week 9/10

[Lusin_Yengibaryan_3B]

So I got this question correct on sapling, the answer was A, but I don't fully understand the concept and why the other answer choices are wrong.

Here is the question:

For the reversible, one‑step reaction,

A+A⥫⥬=k−1k1B+C

the rate constant for the forward reaction, k1 , is 273 L⋅mol−1⋅min−1 and the rate constant for the reverse reaction, k1 , is 399 L⋅mol−1⋅min−1 at a given temperature. The activation energy for the forward reaction is 41.7 kJ⋅mol−1 , whereas the activation energy for the reverse reaction is 20.8 kJ⋅mol−1 .

What effect will raising the temperature of the reaction have on the rate constants and the equilibrium constant?

a)Raising the temperature will increase the forward rate constant, k1 more than it will increase the reverse rate constant, k−1 , resulting in an increase in the equilibrium constant, K .

b)Raising the temperature will increase the forward rate constant, k1 , and reverse rate constant, k−1 , equally, leaving the equilibrium constant, K , unchanged.

c)Raising the temperature will not affect either rate constant or the equilibrium constant, K .

d)Raising the temperature will increase the reverse rate constant, k−1 , more than it will increase the forward rate constant, k1 , resulting in a decrease in the equilibrium constant, K .

[jessicaosuna_1F]

Since the reaction is endothermic, the activation energy for the forward reaction is greater, and so the increase in temp will aid the forward reaction more/ increase k1 more. Since the equilibrium constant, K=k1/k1rev it will lead to an increase in K as well.

B is incorrect on the grounds that they won't be raised equally (dif Activation energy values)

C is incorrect because the temp change clearly has an impact

D is incorrect because it increases K1 more than K1 reverse

[claire ikemiya_2I]

I will try and give an analogy that helps me visualize this. Think about a very wealthy community. Imagine giving each individual $2000. Now think about a much poorer community. Imagine giving each individual $2000. The money will have a much greater impact on the poorer community than the wealthier community.
Now comparing this to the activation energies. Regarding the reaction with a high activation energy, the molecules need more energy to overcome this barrier (analogous to the poor community in need of money). Regarding the reaction with a low activation energy, more of the molecules already have enough energy without the extra energy supplied from temperature (analogous to the extra money) to overcome this barrier. So, increasing the temperature and inputting more energy (giving money) will have a greater effect on the reaction with a higher activation energy. --> Increased temp will increase rate constant of reaction with greater activation energy more than it will increase rate constant of reaction with lower activation energy


This is an explanantion from another person and its really helpful

[Manseej Khatri 2B]

Hi. To add on, if you look at the Arrhenius equation there is the expression Ea/T. The larger the value in the numerator, the larger the difference of that expression when increasing or decreasing the temperature. For example 5/2 and 5/1 has a smaller difference compared to 100/2 and 100/1. Because of the exponential expression, the larger difference in that expression contributes significantly. I should also note that in the equation, -Ea/RT is negated, so the larger the Ea the difference is actually smaller, but because the equilibrium constant K is the ratio of the two rates, it's really the magnitude of difference between the two expressions that matter. For example 10/0.01 = 1000 but 20/0.11 is approximately 181. So even though the change in the numerator (10) is much greater than the change in the denominator (.1), .1 is a magnitude of ten greater than .01 so it has a larger effect.

Another way to look at it is that it is an endothermic reaction and thus takes heat from the environment. The higher the temperature, the greater the heat in the environment and thus the endothermic pathway is more favored.

[Ally M]

Thank you! It's just because it will increase the rate constant of the one with greater activation energy correct?

[Sarah Khan]

To add on: for the first part where you're asked to find K simply just use k1(aka the forward)/k1 (aka the reverse) with the values you are given.