**Using Hess’s Law**

In designing a chemical plant for manufacturing the plastic polyethylene, you need to know the enthalpy change for the removal of H_2 from C_2H_6 (ethane) to give C_2H_4 (ethylene), a key step in the process.

C_2H_6(g) → C_2H_4(g) + H_2(g) ΔH° = ?

From experiments you know these thermochemical expressions:

(a) 2 C_2H_6(g) + 7 O2(g) → 4 CO_2(g) + 6 H_2O(\ell) ΔH^\circ_a = – 3119.4 kJ

(b) C_2H_4(g) + 3 O_2(g) → 2 CO_2(g) ++ 2 H_2O(\ell) ΔH^\circ_b = – 1410.9 kJ

(c) 2 H_2(g) + O_2(g) → 2 H_2O(\ell) ΔH^\circ_c = – 571.66 kJ

Use this information to find the value of ΔH° for the formation of ethylene from ethane.

Step-by-Step

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ΔH° = 137.0 kJ

**Strategy and Explanation ** Analyze reactions (a), (b), and (c). Reaction (a) involves 2 mol ethane on the reactant side, but only 1 mol ethane is required in the desired reaction. Reaction (b) has C_2H_4 as a reactant, but C_2H_4 is a product in the desired reaction. Reaction (c) has 2 mol H_2 as a reactant, but 1 mol H_2 is a product in the desired reaction. First, since the desired expression has only 1 mol ethane on the reactant side, we multiply expression (a) by \frac{1}{2} to give an expression (a’) that also has 1 mol ethane on the reactant side. Halving the coefficients in the equation also halves the enthalpy change.

Next, we reverse expression (b) so that C_2H_4 is on the product side, giving expression (b’). This also reverses the sign of the enthalpy change.

(b') = -(b) 2CO_2(g) + 2 H_2O(\ell)→ C_2H_4(g) + 3 O_2(g) ΔH^\circ_{b'} = – ΔH^\circ_{b} = + 1410.9 kJTo get 1 mol H_2(g) on the product side, we reverse expression (c) and multiply all coefficients by \frac{1}{2} This changes the sign and halves the enthalpy change.

(c') = – \frac{1}{2} (c) H_2O(\ell) → H_2(g) + \frac{1}{2} O_2(g) ΔH^\circ_{c'} = -\frac{1}{2} ΔH^\circ_{c} = + 285.83 kJNow it is possible to add expressions (a’), (b’), and (c’) to give the desired expression.

(a') C_2H_6(g) + \cancel{\frac{7}{2}O_2(g)}→ \cancel{2 CO_2(g)} + \cancel{3 H_2O(\ell)} ΔH^\circ_{a'} = – 1559.7 kJ (b') \cancel{2CO_2(g)} + \cancel{2 H_2O(\ell)}→ C_2H_4(g) + \cancel{3 O_2(g)} ΔH^\circ_{b'} = + 1410.9 kJ \underline{(c') \cancel{H_2O(\ell)} → H_2(g) + \cancel{\frac{1}{2} O_2(g)} ΔH^\circ_{c'} = + 285.83 kJ}Net equation: C_2H_6(g) → C_2H_4(g) + H_2(g) ΔH^\circ_{net} = 137.0 kJ

When the chemical equations are added, there is \frac{7}{2} mol O_2(g) on the reactant side and (3 + \frac{1}{2}) = \frac{7}{2} mol O_2(g) on the product side. There is 3 mol H_2O(\ell) on each side and 2 mol CO_2(g) on each side. Therefore, O_2(g), CO_2(g), and H_2O(\ell) all cancel, and the chemical equation for the conversion of ethane to ethylene and hydrogen remains.

**Reasonable Answer Check** The overall process involves breaking a molecule apart into simpler molecules, which is likely to involve breaking bonds. Therefore it should be endothermic, and ΔH° should be positive.

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