Propane, C_3H_6(g), reacts with HCl(g) to produce 2-chloropropane, C_3H_7Cl(g) a
ID: 517349 • Letter: P
Question
Propane, C_3H_6(g), reacts with HCl(g) to produce 2-chloropropane, C_3H_7Cl(g) as illustrated below: C_3H_6(g) + HCl rightarrow C_3H_7 Cl(g) The relevant bond energies, in k|/mol are: C-C 348 C=C 614 C-H 413 H-Cl 431 C-Cl 328 A. Find Delta H degree for this reaction at 298 K. B. Given the standard molar entropies, as follows. i .Find Delta S degree. Include units. ii. Find Delta G degree. Include units. iii. Find the equilibrium constant, K_p, for this reaction. C. If the pressures of all three gases in the reaction mixture were 0.50 atm., would the reaction shift so as to produce more or less 2-chloropropane? Perform a calculation that supports your answer. D. In fact, at standard pressure and 298 K, 2-chtoropropane is a liquid. For the reaction C_3H_6(g) + HCl(g) rightarrow C_3H_7Cl(l) i Would Delta S degree be larger or smaller than the Delta S degree value calculated in part B i? Explain ii. Would Delta H degree be larger or smaller than the Delta H degree value calculated in part A? Explain.Explanation / Answer
(A) deltaH0 = 6 deltaH0b(C-H)+deltaH0b(C=C)+deltaH0b(C-C)+deltaH0b(H-Cl)-7deltaH0b(C-H)-2deltaH0b(C-C)-deltaH0b(C-Cl)
deltaH0b = 6(413)+614+348+431-7(413)-2(348)-328
deltaH0 = - 44 kJ / mol
(B) (i)
deltaS0 = deltaS0f(C3H7Cl) - deltaS0f(C3H6) - deltaS0f(HCl)
= 304 - 267 - 187
= - 150 J/mol.K
= - 0.150 kJ/mol.K
(ii) deltaG0 = deltaH0 - TdeltaS0
= - 44 - 298(-0.150)
= + 0.700 kJ/mol
deltaG0 = - R T lnKp
0.700 = - 0.008314 * 298 * lnKp
Kp = e-0.282
Kp = 0.754
(C) Qp = 0.50 / (0.50 * 0.50) = 2
Since Qp > Kp , the reactions moves towards left side i.e reactants side.
(D)
(i) Since liquids have lower entropy, net entropy cahnge is low.
(ii) Druing concersion of gas to liquid it loses some energy. SO, enthalpy change is more negative.
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