# Question 6.PS.7: Thermochemical Expressions Given the thermochemical expressi......

Thermochemical Expressions

Given the thermochemical expression

$2 C_2H_6(g) + 7 O_2(g) → 4 CO_2(g) + 6 H_2O(g) ΔH° = – 2856 kJ$

write a thermochemical expression for
(a) Formation of 1 mol $CO_2(g)$ by burning $C_2H_6(g)$
(b) Formation of 1 mol $C_2H_6(g)$ by reacting $CO_2(g)$ with $H_2O(g)$
(c) Combination of 1 mol $O_2(g)$ with a stoichiometric quantity of $C_2H_6(g)$

Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.

(a)  $\frac{1}{2}C_2H_6(g) + \frac{7}{4}O_2(g) → CO_2(g) + \frac{3}{2}H_2(g) ΔH° = – 714.0 kJ$

(b)  $2 CO_2(g) + 3 H_2O(g) → C_2H_6(g) + \frac{7}{4}O_2(g) ΔH° = +1428 kJ$

(c))  $\frac{2}{7}C_2H_6(g) + O_2(g) → \frac{4}{7}CO_2(g) + \frac{6}{7}H_2(g) ΔH° = – 408.0 kJ$

Strategy and Explanation
(a) Producing 1 mol $CO_2(g)$ requires that one quarter the molar amount of each reactant and product be used and also makes the $ΔH°$ value one quarter as big.
(b) Forming $C_2H_6(g)$ means that $C_2H_6(g)$ must be a product. This changes the direction of the reaction and the sign of $ΔH°$; forming 1 mol $C_2H_6(g)$ requires that each coefficient be halved and this halves the size of $ΔH°$.
(c) If 1 mol $O_2(g)$ reacts, only mol $\frac{2}{7} C_2H_6(g)$ is required; each coefficient is one seventh its original value, and $ΔH°$ is also one seventh the original value.

Reasonable Answer Check   In each case examine the coefficients, the direction of the chemical equation, and the sign of $ΔH°$ to make certain that the appropriate quantity of reactant or product and the appropriate sign have been written.

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