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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
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(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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