Question

energy changes

1. ethene gas, ( c_{2}h_{4(g)} ), is the raw material for the synthesis of the plastic polyethylene. use hesss law to help the engineers (who are designing a process to make ethene from ethane gas, ( c_{2}h_{6(g)} )) determine the change in enthalpy (( delta h^{circ} )) of the desired balanced reaction below, given the following equations and their enthalpy changes.

( c_{2}h_{6(g)} \to c_{2}h_{4(g)}+h_{2(g)} )

the engineers have the following thermochemical equations:

( c_{2}h_{6(g)}+\frac{7}{2}o_{2(g)} \to 2co_{2(g)}+3h_{2}o_{(l)} quad delta h^{circ}=-1559 mathrm{~kj} )

( c_{2}h_{4(g)}+3o_{2(g)} \to 2co_{2(g)}+2h_{2}o_{(l)} quad delta h^{circ}=-1411 mathrm{~kj} )

( 2h_{2(g)}+o_{2(g)} \to 2h_{2}o_{(l)} quad delta h^{circ}=-572 mathrm{~kj} )

Explanation:

Step1: Identify Target Reaction

Target: $\ce{C2H6(g) -> C2H4(g) + H2(g)}$
Given reactions:

1. $\ce{C2H6(g) + 7/2 O2(g) -> 2 CO2(g) + 3 H2O(l)}$ $\Delta H^\circ = -1559\ \text{kJ}$
2. $\ce{C2H4(g) + 3 O2(g) -> 2 CO2(g) + 2 H2O(l)}$ $\Delta H^\circ = -1411\ \text{kJ}$
3. $\ce{2 H2(g) + O2(g) -> 2 H2O(l)}$ $\Delta H^\circ = -572\ \text{kJ}$

Step2: Manipulate Reactions

• Reaction 1: Keep as is (produces $\ce{C2H6}$ reactant, $\ce{CO2, H2O}$ products).
• Reaction 2: Reverse (to get $\ce{C2H4}$ product, $\ce{CO2, H2O}$ reactants). New $\Delta H^\circ = +1411\ \text{kJ}$ (reverse flips sign).
• Reaction 3: Reverse and halve (to get $\ce{H2}$ product, $\ce{H2O}$ reactant). Reverse: $\ce{2 H2O(l) -> 2 H2(g) + O2(g)}$ $\Delta H^\circ = +572\ \text{kJ}$; halve: $\ce{H2O(l) -> H2(g) + 1/2 O2(g)}$ $\Delta H^\circ = +286\ \text{kJ}$.

Step3: Combine Reactions

Add modified reactions:

1. $\ce{C2H6(g) + 7/2 O2(g) -> 2 CO2(g) + 3 H2O(l)}$ $\Delta H_1 = -1559\ \text{kJ}$
2. $\ce{2 CO2(g) + 2 H2O(l) -> C2H4(g) + 3 O2(g)}$ $\Delta H_2 = +1411\ \text{kJ}$
3. $\ce{H2O(l) -> H2(g) + 1/2 O2(g)}$ $\Delta H_3 = +286\ \text{kJ}$

Sum reactants/products:
$\ce{C2H6(g) + 7/2 O2 + 2 CO2 + 3 H2O -> 2 CO2 + 3 H2O + C2H4 + 3 O2 + H2 + 1/2 O2}$

Simplify (cancel $\ce{2 CO2, 2 H2O, 3 O2 + 1/2 O2 - 7/2 O2 = 0}$):
$\ce{C2H6(g) -> C2H4(g) + H2(g)}$ (matches target).

Step4: Sum Enthalpies

$\Delta H^\circ_{\text{target}} = \Delta H_1 + \Delta H_2 + \Delta H_3$
$= -1559 + 1411 + 286$
$= (-1559 + 1411) + 286$
$= (-148) + 286$
$= 138\ \text{kJ}$

Answer:

The enthalpy change $\Delta H^\circ$ for the reaction is $\boldsymbol{+138\ \text{kJ}}$ (or $138\ \text{kJ}$).