Question

The heat combustion of glucose (C6H12O6) is desrcribed by the process:

C6H12O6 + 6 O2 <-> 6 CO2 + 6 H2O

for which it is known that delta H = -2816 kJ/mol

a. If the temperature is increased, what effect does this have on the reaction mixture: a shift toward reactants or a shift toward products and why?

b. Temperature does change the value of the equilibrium constant K. In this example, how is K affected on increasing temperature?

Answer 1

a) Equilibrium will shift to the left i.e. reactants

b) K will increase with increase in temperature

Step-by-step explanation:

The given reaction is:

C6H12O6 + 6 O2 ↔ 6 CO2 + 6 H2O ΔH = -2816 kJ/mol

a) As per Le Chatelier's principle, for a reaction at equilibrium any changes in temperature, pressure or concentration will shift the equilibrium in a direction so as to undo the effect of the induced change.

The given reaction is exothermic (since ΔH is negative) i.e. it accompanied by the release of heat and hence an increase in temperature. Therefore, if the temperature is increased the equilibrium will shift in the opposite direction i.e. towards the left or towards the reactants.

b) The equilibrium constant (K) and temperature T are related via the Van't Hoff equation:

`[ln(K2)/(K1)=(-\Delta H)/(R)[(1)/(T2)-(1)/(T1)]`

In the given reaction, ΔH is negative and the condition is T2>T1

Therefore, K2 > K1

The value of equilibrium constant will increase with increase in temperature