Question

Date 02/12/2020 PreLab Assignment Experiment 3: HESS is Just a Four Letter Word Consider the following reactions 1 CHaq) 2 Org) - CO2(g) + 2 H2O(l): AH=-890 kJ/mol 2 2 CHelg) + 4 Ong) -- 2 CO2(g) + 4H2O(l): AH=-1780 kJ/mol 3. CO2(g) + 2 H2O(l) - CHI(g) + 2 O2(g): AH= 890 kJ/mol 2 H2O(l) -- 2 H2O(g): AH = 88 kJ/mol Quantitatively, how does the change in enthalpy, AH for reaction #1 compare to the change in enthalpy for reaction #2? How would you expect the heat released by the combustion of 1.0 g of CH4 to compare to the heat released by the combustion of 2.00 g CH4 ? Based upon your answers to the 2 questions above would you conclude that AH is an extensive or intensive property? Explain. Quantitatively, how does the value of AH for reaction #1 compare to the AH for reaction #3? Based upon your answer to the above question what can you say in general about the AH for a forward reaction compared to the AH for a reverse reaction?

Answer 1

AS PER CHEGG GUIDELINES, I HAVE ATTEMPTED THE FIRST QUESTION ONLY AS NO SPECIFIC QUESTION WAS ASKED TO SOLVE.

Reaction 1 - $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l),\: \: \Delta H=-890 kJ/mol$

Reaction 2 - $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l),\: \: \Delta H=-1780 kJ/mol$

We can see that on doubling the number of moles of reactants, the enthalpy of the reaction is also doubling. This is because enthalpy is an extensive property, i.e. it depends on the amount of substance. For example, mass is an extensive property as more the amount of material, higher is the mass. On the other hand, temperature is an intensive property as it does not depend on the amount of material.

The expression for the change in enthalpy is given by-

  $\Delta H=\Delta U+P\Delta V$   

Here, $\Delta U$ is extensive ( more the amount of material, greater the internal energy)

P is intensive, but $\Delta V$ is again extensive ( more the amount of material, greater the internal energy). So the product $P\Delta V$ is again extensive.

So $\Delta H$ is extensive.

As the amount of reactants and products is doubled in reaction 2, so $\Delta H$ is also exactly doubled.