Calculate H0rxn for each of the following using data from Appendix B.
(a) 2 H2S(g) + 3 O2(g) 2 SO2(g) + 2 H2O(g)
b) CH4(g) + Cl2(g) CCl4(l) + HCl(g) [unbalanced]
To calculate H0rxn (standard enthalpy change) for each of the given reactions, you need to know the standard enthalpy of formation (ΔHf°) values of the involved compounds. The ΔHf° values for most compounds can be found in Appendix B of a chemistry textbook or through reliable online sources.
Let's calculate H0rxn for each reaction.
(a) 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)
The reaction can be balanced already.
To calculate H0rxn, you need the ΔHf° values of the reactants and the products. Look up the ΔHf° values for H2S(g), O2(g), SO2(g), and H2O(g) in Appendix B or an online source.
Assuming the ΔHf° values are as follows:
ΔHf°(H2S) = -20 kJ/mol
ΔHf°(O2) = 0 kJ/mol
ΔHf°(SO2) = -300 kJ/mol
ΔHf°(H2O) = -240 kJ/mol
Now, use the formula:
H0rxn = (Σ ΔHf° products) - (Σ ΔHf° reactants)
H0rxn = (2 * ΔHf°(SO2)) + (2 * ΔHf°(H2O)) - (2 * ΔHf°(H2S)) - (3 * ΔHf°(O2))
Substitute the values and calculate:
H0rxn = (2 * -300 kJ/mol) + (2 * -240 kJ/mol) - (2 * -20 kJ/mol) - (3 * 0 kJ/mol)
H0rxn = -600 kJ/mol - 480 kJ/mol + 40 kJ/mol
H0rxn = -1040 kJ/mol
Therefore, H0rxn for the reaction 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) is -1040 kJ/mol.
(b) CH4(g) + Cl2(g) → CCl4(l) + HCl(g) [unbalanced]
This reaction is unbalanced, so the first step is to balance the equation:
CH4(g) + 4 Cl2(g) → CCl4(l) + 4 HCl(g)
Once you have the balanced equation, you can proceed to calculate H0rxn using the same method as in the previous example. Find the ΔHf° values for CH4(g), Cl2(g), CCl4(l), and HCl(g) in Appendix B or an online source.
Assuming the ΔHf° values are as follows:
ΔHf°(CH4) = -74.8 kJ/mol
ΔHf°(Cl2) = 0 kJ/mol
ΔHf°(CCl4) = -95.7 kJ/mol
ΔHf°(HCl) = -92.3 kJ/mol
Now apply the formula:
H0rxn = (Σ ΔHf° products) - (Σ ΔHf° reactants)
H0rxn = ΔHf°(CCl4) + (4 * ΔHf°(HCl)) - ΔHf°(CH4) - (4 * ΔHf°(Cl2))
Substitute the values and calculate:
H0rxn = -95.7 kJ/mol + (4 * -92.3 kJ/mol) - (-74.8 kJ/mol) - (4 * 0 kJ/mol)
H0rxn = -95.7 kJ/mol - 369.2 kJ/mol + 74.8 kJ/mol
H0rxn = -390.1 kJ/mol
Therefore, H0rxn for the reaction CH4(g) + Cl2(g) → CCl4(l) + HCl(g) is -390.1 kJ/mol.
Remember, the accuracy of the calculated H0rxn values depends on the accuracy of the ΔHf° values used. Always refer to reliable sources for accurate ΔHf° data.