Calculate the following quantities. Refer to the standard entropy values as needed.

Question

Consider the following reaction at 298 K.

C(graphite)+2H2(g)⟶CH4(g) Δ𝐻∘=−74.6 kJ

ΔS_sys= J/K

ΔS_surr= J/K

ΔS_univ= J/K

Answer 1

1 mole C(graphite), 2 moles H2(g), and 1 mole CH4(g)

The standard enthalpy of formation of each molecule are as follows:

C(graphite): 0 kJ/mol because graphite is carbon’s “reference state,” and reference states have standard enthalpy of 0 kJ/mol.

H2(g): 0 kJ/mol because H2(g) is the “reference state” of hydrogen

CH4(g): -74.2 – -74.9 kJ/mol according to google. I’m going to assume it is -74.6 kJ/mol.

The formula for calculating the enthalpy of a reaction is the enthalpy of the products minus the enthalpy of the reactants. Since both reactants in this problem have an enthalpy of 0 kJ/mol, the enthalpy of this reaction equals the enthalpy of the product. Thus, the reaction’s enthalpy of -74.6 kJ matches the enthalpy of CH4(g) of -74.6 kJ/mol.

Since the enthalpy of CH4(g) is “per mole” as shown by “/mol”, we can say that there is 1 mole of CH4(g) in reaction.

From there, we can use the equation’s molar ratio of 1 C(graphite): 2 H2(g) : 1 CH4(g) to get the following amounts of each molecule —

1 mole of C(graphite)

2 moles of H2(g)

1 mole of CH4(g)

NOTE: If the enthalpy of CH4(g) didn’t exactly match the enthalpy of the reaction, then you would divide the enthalphy of CH4(g) by the enthalpy of the reaction to get the amount of moles of CH4(g). From there, as we did above, you use the molar ratios in the equation to find the mole amounts of the other molecules.

For example, if the standard enthalpy of formation of CH4(g) was -150 kJ/mol, the amount of moles of CH4(g) would equal -150/-74.6=2.011.

If the problem wants integer values, you would round to the closest integer, which in this case would be 2 moles. From there, mole ratios indicate that there would be 2 moles of C(graphite) and 4 moles of H2(g).