Question

Parts A and B: Calculate ΔrS0298 and ΔrH0298 for the combustion of hydrogen gas with oxygen gas to yield 0.70 moles of water vapor.

Part C: Calculate (ΔrH0350−ΔrH0298) for the combustion of hydrogen gas with oxygen gas to yield 0.70 moles of water vapor. The molar heat capacities at constant pressure for H2(g), O2(g) and H2O are 28.8 J.K-1.mol-1, 29.4 J.K-1.mol-1 and 36.5 K-1.mol-1, respectively.

Answer 1

To answer the student's question, one must use standard thermodynamic tables for Part A and B to find ΔrS°298 and ΔrH°298, and Kirchhoff's law in Part C to calculate the difference in enthalpy changes (ΔrH°350 - ΔrH°298) for the combustion of hydrogen gas.

Step-by-step explanation:

The combustion of hydrogen gas with oxygen gas to yield water vapor can be represented by the balanced chemical equation:

H2(g) + 1/2 O2(g) → H2O(g)

For Part A and B, we need to calculate the standard entropy change (ΔrS°298) and the standard enthalpy change (ΔrH°298) for this reaction. Using standard thermodynamic tables, we would look up the standard molar entropies and enthalpies of formation for the reactants and products.

Part C involves calculating the enthalpy change at 350 K (ΔrH°350) compared to 298 K (ΔrH°298) for the same reaction. To do this, we would apply Kirchhoff's law, which relates the heat capacities of the reactants and products to the change in enthalpy at different temperatures.

The student's question mentions specific heat capacities for H2(g), O2(g), and H2O(g). These would be used in Kirchhoff's law equation to find the enthalpy changes between the two temperatures for the given moles of water vapor produced.