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Question: Heat of combustion of $C_{(s)}$, $H_{2(g)}$ and $C_{2}H_{6(g)}$ are $-x_1$, $-x_2$ and $-x_3$ respec...

Heat of combustion of C(s)C_{(s)}, H2(g)H_{2(g)} and C2H6(g)C_{2}H_{6(g)} are −x1-x_1, −x2-x_2 and −x3-x_3 respectively. Hence heat of formation of C2H6(g)C_2H_{6(g)} is

A

-2x_1 - 3x_2 + x_3

B

-x_1 - x_2 + x_3

C

x_1 + x_2 - x_3

D

-x_3 + 2x_1 + 3x_2

Answer

-2x_1 - 3x_2 + x_3

Explanation

Solution

Write the combustion reaction of ethane:

C2H6+72O2→2CO2+3H2OΔH=−x3C_2H_6 + \frac{7}{2}O_2 \to 2CO_2 + 3H_2O \quad \Delta H = -x_3.

For the combustion of the elements:

C(s)+O2→CO2ΔH=−x1C(s) + O_2 \to CO_2 \quad \Delta H = -x_1,

H2(g)+12O2→H2OΔH=−x2H_2(g) + \frac{1}{2}O_2 \to H_2O \quad \Delta H = -x_2.

Thus, formation of combustion products from elements gives:

2CO2:2C(s)+2O2→2CO2ΔH=−2x12CO_2: \quad 2C(s) + 2O_2 \to 2CO_2 \quad \Delta H = -2x_1,

3H2O:3H2(g)+32O2→3H2OΔH=−3x23H_2O: \quad 3H_2(g) + \frac{3}{2}O_2 \to 3H_2O \quad \Delta H = -3x_2.

Combining, formation of the products:

2C(s)+3H2(g)+(2+32)O2→2CO2+3H2OΔH=−2x1−3x22C(s) + 3H_2(g) + \left(2+\frac{3}{2}\right)O_2 \to 2CO_2 + 3H_2O \quad \Delta H = -2x_1 - 3x_2.

Now think of the combustion of ethane as the difference between the formation of products and the formation of ethane (from its elements):

ΔHcomb(C2H6)=[formation of products]−[ΔHf(C2H6)]\Delta H_{\text{comb}} (C_2H_6) = [\text{formation of products}] - [\Delta H_f (C_2H_6)].

That is,

−x3=(−2x1−3x2)−ΔHf(C2H6)-x_3 = (-2x_1 - 3x_2) - \Delta H_f (C_2H_6).

Rearrange to solve for the heat of formation:

ΔHf(C2H6)=−2x1−3x2+x3\Delta H_f (C_2H_6) = -2x_1 - 3x_2 + x_3.