Question

a chemist measures the enthalpy change \\( \delta h \\) during the following reaction:

\\( \mathrm { c } _ { 6 } \mathrm { h } _ { 6 } ( l ) + 3 \mathrm { h } _ { 2 } ( g ) \
ightarrow 3 \mathrm { c } _ { 2 } \mathrm { h } _ { 4 } ( g ) \quad \delta h = 108. \mathrm { kj } \\)

use this information to complete the table below. round each of your answers to the nearest \\( \mathrm { kj } \\).

Explanation:

Step1: Reverse the reaction

When a reaction is reversed, the sign of $\Delta H$ changes.
For the reaction \(3C_{2}H_{4}(g)\to C_{6}H_{6}(l)+3H_{2}(g)\), since it is the reverse of \(C_{6}H_{6}(l)+3H_{2}(g)\to 3C_{2}H_{4}(g)\) (\(\Delta H = 108\space kJ\)), \(\Delta H=- 108\space kJ\)

Step2: Scale the reaction

When a reaction is multiplied by a factor \(n\), \(\Delta H\) is also multiplied by \(n\).
For the reaction \(15C_{2}H_{4}(g)\to 5C_{6}H_{6}(l)+15H_{2}(g)\), the original reversed reaction \(3C_{2}H_{4}(g)\to C_{6}H_{6}(l)+3H_{2}(g)\) (\(\Delta H=- 108\space kJ\)) is multiplied by \(5\). So \(\Delta H = 5\times(-108)=-540\space kJ\)

Step3: Scale the original reaction

For the reaction \(3C_{6}H_{6}(l)+9H_{2}(g)\to 9C_{2}H_{4}(g)\), the original reaction \(C_{6}H_{6}(l)+3H_{2}(g)\to 3C_{2}H_{4}(g)\) (\(\Delta H = 108\space kJ\)) is multiplied by \(3\). So \(\Delta H=3\times108 = 324\space kJ\)

Answer:

• \( - 108\space kJ\)
• \( - 540\space kJ\)
• \(324\space kJ\)