Question

using bond dissociation energies, determine the approximate enthalpy of reaction for the following reaction:
2 h₂s (g) + 3 o₂ (g) → 2 so₂ (g) + 2h₂o (g)
note that bond dissociation energies can be found in the tables of chemical data posted on canvas. report the answer in kilojoules to 3 sig figs, but do not include units in your answer.

Explanation:

Step1: Identify Bonds Broken/Formed

Bonds broken: In \(2\ce{H2S}\), there are \(2\times2 = 4\) \(\ce{H-S}\) bonds and \(3\) \(\ce{O=O}\) bonds.
Bonds formed: In \(2\ce{SO2}\), there are \(2\times2 = 4\) \(\ce{S=O}\) bonds (each \(\ce{SO2}\) has two double bonds), and in \(2\ce{H2O}\), there are \(2\times2 = 4\) \(\ce{O-H}\) bonds.

Step2: Recall Bond Dissociation Energies (BDE)

Typical BDE values (from tables):
- \(\ce{H-S}\): ~366 kJ/mol
- \(\ce{O=O}\): ~498 kJ/mol
- \(\ce{S=O}\) (in \(\ce{SO2}\)): ~522 kJ/mol (double bond)
- \(\ce{O-H}\): ~464 kJ/mol

Step3: Calculate Energy for Bonds Broken

Energy to break bonds:
\(4\times366 + 3\times498 = 1464 + 1494 = 2958\) kJ (for 2 mol \(\ce{H2S}\) and 3 mol \(\ce{O2}\))

Step4: Calculate Energy for Bonds Formed

Energy released from forming bonds:
\(4\times522 + 4\times464 = 2088 + 1856 = 3944\) kJ (for 2 mol \(\ce{SO2}\) and 2 mol \(\ce{H2O}\))

Step5: Enthalpy of Reaction (\(\Delta H\))

\(\Delta H = \text{Bonds Broken} - \text{Bonds Formed} = 2958 - 3944 = -986\) kJ (per reaction, considering stoichiometry). Wait, let's recheck stoichiometry:

Wait, correct stoichiometry:
- \(2\ce{H2S}\): 2 molecules, each with 2 \(\ce{H-S}\) → 4 \(\ce{H-S}\)
- \(3\ce{O2}\): 3 \(\ce{O=O}\)
- \(2\ce{SO2}\): each \(\ce{SO2}\) has two \(\ce{S=O}\) (double bonds) → 2×2 = 4 \(\ce{S=O}\)
- \(2\ce{H2O}\): each \(\ce{H2O}\) has two \(\ce{O-H}\) → 2×2 = 4 \(\ce{O-H}\)

Wait, maybe BDE values differ. Let's use standard values:
\(\ce{H-S}\): 366 kJ/mol, \(\ce{O=O}\): 498 kJ/mol, \(\ce{S=O}\) (in \(\ce{SO2}\)): 522 kJ/mol (average for double bond), \(\ce{O-H}\): 464 kJ/mol.

Bonds broken:
\(2\times(2\times366) + 3\times498 = 2\times732 + 1494 = 1464 + 1494 = 2958\) kJ

Bonds formed:
\(2\times(2\times522) + 2\times(2\times464) = 2\times1044 + 2\times928 = 2088 + 1856 = 3944\) kJ

\(\Delta H = 2958 - 3944 = -986\). But wait, maybe the \(\ce{S=O}\) bond in \(\ce{SO2}\) is actually two double bonds? Wait, \(\ce{SO2}\) has a resonance structure with two S=O bonds (each ~522 kJ/mol). Alternatively, maybe I made a mistake in stoichiometry. Let's check the reaction again: \(2\ce{H2S} + 3\ce{O2} → 2\ce{SO2} + 2\ce{H2O}\).

Wait, another approach: Enthalpy of reaction = sum of BDE of reactants - sum of BDE of products.

Reactants: \(2\ce{H2S}\) (4 H-S) and \(3\ce{O2}\) (3 O=O).
Products: \(2\ce{SO2}\) (4 S=O) and \(2\ce{H2O}\) (4 O-H).

So:
Reactants BDE: \(4\times366 + 3\times498 = 1464 + 1494 = 2958\)
Products BDE: \(4\times522 + 4\times464 = 2088 + 1856 = 3944\)
\(\Delta H = 2958 - 3944 = -986\). Rounding to 3 sig figs, it's -986 (or maybe -987? Wait, let's check precise BDE values.

Wait, actual BDE values (more accurate):
- \(\ce{H-S}\): 366 kJ/mol
- \(\ce{O=O}\): 498 kJ/mol
- \(\ce{S=O}\) (in \(\ce{SO2}\)): 522 kJ/mol (average bond energy for S=O in \(\ce{SO2}\))
- \(\ce{O-H}\): 464 kJ/mol

Yes, so the calculation gives -986, which is -986 (3 sig figs). Wait, but maybe the problem uses different BDE values. For example, if \(\ce{S=O}\) is 552 kJ/mol? Let's recalculate:

If \(\ce{S=O}\) is 552:
Products BDE: \(4\times552 + 4\times464 = 2208 + 1856 = 4064\)
\(\Delta H = 2958 - 4064 = -1106\) → no, that's not. Wait, maybe I messed up the number of bonds. Let's count again:

- \(\ce{H2S}\): H-S-H, so 2 H-S bonds per molecule. 2 molecules: 4 H-S.
- \(\ce{O2}\): O=O, 3 molecules: 3 O=O.
- \(\ce{SO2}\): S=O (double bond) and S-O (single bond)? No, \(\ce{SO2}\) has two S=O bonds (resonance), so each \(\ce{SO2}\) has two S=O bonds. So 2 \(\ce{SO2}\): 4 S=O.
- \(\ce{H2O}\): H-O-H, 2 O-H bonds per molecule. 2 molecules: 4 O-H.

Yes, that's correct. So using the standard BDE values, the enthalpy is -986, which to 3 sig figs is -986 (or maybe -987 if rounding). Wait, maybe the actual BDE values from the table are different. Let's assume the table has:

\(\ce{H-S}\): 368 kJ/mol, \(\ce{O=O}\): 495 kJ/mol, \(\ce{S=O}\) (in \(\ce{SO2}\)): 523 kJ/mol, \(\ce{O-H}\): 463 kJ/mol.

Then:

Bonds broken: \(4\times368 + 3\times495 = 1472 + 1485 = 2957\)
Bonds formed: \(4\times523 + 4\times463 = 2092 + 1852 = 3944\)
\(\Delta H = 2957 - 3944 = -987\).

So the answer is approximately -986 or -987. Depending on the table, but the standard calculation gives around -986 to -987.

Answer:

-986