1.Hess's Law ProblemsCalculate the AH for the reaction CHa (g) + Ha (g) - C»He (g), from the following:a.CHa (g) + 3 0, (g) › 2 CO, (g) + 2 H,0 (1)2 C_Ho (g) + 7 02 (g) - 4 COz (g) + 6 H,0 (1)C.2 H2 (g) + 02 (g)-2 H,0 (1)IAH = - 1411. kJAH = - 3120. kJAH = - 571.6 kJ

[tex]\begin{gathered} a.\text{ }C_2H_4(g)+3O_2(g)\rightarrow2CO_2(g)+2H_2O(l)\times2\text{ }\Delta H=-1411\text{ }kJ\times2 \\ \\ a.\text{ }2C_2H_4(g)+6O_2(g)\rightarrow4CO_2(g)+4H_2O(l)\text{ }\Delta H=-2822\text{ }kJ \end{gathered}[/tex]

Step 2: Reverse (b)

[tex]\begin{gathered} b.\text{ }2C_2H_6(g)+7O_2(g)\rightarrow4CO_2(g)+6H_2O(l)\text{ }\Delta H=-3120\text{ }kJ \\ \\ Reverse\text{ }the\text{ }equation \\ \\ b.\text{ }4CO_2\left(g\right)+6H_2O\left(l\right)\rightarrow2C_2H_6\left(g\right)+7O_2\left(g\right)\text{ }\Delta H=+3120\text{ }kJ \end{gathered}[/tex]

Step 3: (c)

[tex]c.\text{ }2H_2(g)+O_2(g)\rightarrow2H_2O(l)\text{ }\Delta H=-571.6\text{ }kJ[/tex]

Step 4: Combine (a), (b), and (c) and simplify.

[tex]\begin{gathered} 2C_2H_4(g)+6O_2(g)+4CO_2\left(g\right)+6H_2O\left(l\right)+2H_2(g)+O_2(g)\rightarrow \\ \\ 4CO_2(g)+4H_2O(l)+2C_2H_6\left(g\right)+7O_2\left(g\right)+2H_2O(l) \\ \\ \Delta H=(-2822\text{ }kJ+3120\text{ }kJ-571.6\text{ }kJ)-(-3120\text{ }kJ+2822\text{ }kJ) \\ \\ Simplify\text{ }the\text{ }equation \\ \\ 2C_2H_4(g)+2H_2(g)\rightarrow2C_2H_6\left(g\right)\text{ }\Delta H=(-273.6\text{ }kJ)-(-298\text{ }kJ) \\ \\ 2C_2H_4(g)+2H_2(g)\rightarrow2C_2H_6\left(g\right)\text{ }\Delta H=+24.4\text{ }kJ \end{gathered}[/tex]

Since the given equation is in 1 mole, then divide through by 2

[tex]\begin{gathered} \frac{2}{2}C_2H_4(g)+\frac{2}{2}H_2(g)\operatorname{\rightarrow}\frac{2}{x}C_2H_6(g)\text{ }\Delta H=\frac{+24.4}{2}\text{ }kJ \\ \\ C_2H_4(g)+H_2(g)\operatorname{\rightarrow}C_2H_6(g)\text{ }\Delta H=+12.2\text{ }kJ \end{gathered}[/tex]

Therefore, ΔH for the given reaction is +12.2 kJ