Question

The balanced chemical equation below shows the combustion of methanol: CH3OH(l) + 3O2(g) → 2 H2O(g) + CO2(g)

a) Predict the sign of ΔS for this reaction and briefly explain your reasoning

b) Predict the sign of ΔH for this reaction and briefly explain your reasoning

c) Is the sign of ΔG temperature dependent in this reaction? Briefly explain your reasoning

Answer 1

a] The sign of $\Delta$ S in the process is positive.

Reasoning: Here in the combustion precess we can see that one of the reactant is liquid (i.e CH3OH) and one is gas(i.e. O2). But on the product side bothe the products are in gas phase. We know that S(solid)<S(liquid)<S(gas) .Thus we can say that the entropy is increasing during the process. Hence $\Delta$ S is positive.

b] The sign of $\Delta$ H is negative.

Reasoning: The combustion process itself is an exothermic process. Because when a substance undergo combustion it always releses energy to the surrounding. Again from the reaction we can see that the number of bond broken during the process is 5 single bond and 3 double bond. However the number of bond formed is two single bond and two double bond. That means number of bond broken during the process is grater than number of bond formed. However the energy required to break bond is always less than the enegry relesed during the formation bond. Hence $\Delta$ H is negative.

In principle if Number of bond broken > Number of bond formed then $\Delta$ H is negative.

c] The sign of $\Delta$ G is  independ on the temperature.

Reasoning: We know that $\Delta$ G = $\Delta$ H - T$\Delta$S

Now since in this process $\Delta$ H is negative and $\Delta$ S is positive

Therefor $\Delta$ G will be always negative irrespective of the value of temperature.