# Question 6.PS.11: Using Hess’s Law In designing a chemical plant for manufactu......

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
The 'Blue Check Mark' means that this solution was answered by an expert.
$ΔH° = 137.0 kJ$

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.

$(a') = \frac{1}{2} (a) C_2H_6(g) + \frac{7}{2}O_2(g)→ 2 CO_2(g) + 3 H_2O(\ell) ΔH^\circ_{a'} = – 1559.7 kJ$

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.

$(b') = -(b) 2CO_2(g) + 2 H_2O(\ell)→ C_2H_4(g) + 3 O_2(g) ΔH^\circ_{b'} = – ΔH^\circ_{b} = + 1410.9 kJ$

To 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.

$(c') = – \frac{1}{2} (c) H_2O(\ell) → H_2(g) + \frac{1}{2} O_2(g) ΔH^\circ_{c'} = -\frac{1}{2} ΔH^\circ_{c} = + 285.83 kJ$

Now 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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