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)
(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.