Consider this system at equilibrium. A (aq) - ⇀ ↽ - B (aq) Δ H = + 850 kJ/mol What can be said about Q and K immediately after an increase in temperature?

Q will increase,

R will remain the same.

Explanation:

We can deduce the equilibrium

A (aq) ⇒ B (aq)

is an endothermic process since ΔH is positive, for which

According to LeChatelier's principle the disturbance of an increase in temperature for the system will favor the reaction that consumes the added heat so as to minimize the change in temperature.

The system initially then produces more B (aq) and Q > K, but the system will relax back, after the added heat is removed, making Q = Keq again.

Therefore inmediately after the increase in temperature Q will be greater than K eliminating the perturbance of the increase in temperature producing more B (aq), but the system will return to equilibrium by making Q = K

Q is greater than K, since the equilibrium constant, K decreased.

Explanation:

In order to be able to solve this equation very well we have to be able to understand the Le Chatliers principle. Therefore, according to the Le Chatliers principle; the equilibrium of a system will shift to counteract the effect of the changes in pressure, temperature, or concentration of a reactant.

Note that Q is the reaction quotent and K is the equilibrium constant.

From the question, we are given a negative value for our ∆H which suggest the reaction to be an exothermic Reaction. In an exothermic reaction, whenever the temperature is increased the equilibrium constant will decrease thereby making the equilibrium to shift to the left side, that is towards the Reactant (s) side.