H\(_3\)PO\(_4\)(aq) + H\(_2\)O(l) ↔ H\(_2\)PO\(^{4-}\)(aq) + H\(_3\)O\(^+\)(aq) ΔH = -13kJ at 298K
In the reaction above, an increase in temperature would
- slow down the reverse reaction
- have no effect on the reaction
-
favour the reverse reaction
- favour the forward reaction
The correct answer is: C
Explanation
In an exothermic reaction, an increase in temperature will favour the reactants (shift the reaction towards the left), meaning the reaction will produce less product as heat is considered a product in an exothermic reaction; according to Le Chatelier's principle, the system will try to counteract the added heat by favouring the reactant side. Thus, the reverse reaction will be favoured.