Holooly Plus Logo

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.
Learn more on how do we answer questions.
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.

Related Answered Questions