-13, -13 2CIF, (g) 39. CIF can be prepared by the following reaction. Ch(g) + 3F
ID: 1043717 • Letter: #
Question
-13, -13 2CIF, (g) 39. CIF can be prepared by the following reaction. Ch(g) + 3F2(g) For CIF, the standard enthalpy change of formation, ??. is-163 kl/mol and the standard free energy change of formation, AGot, is -123 kJ/mol. (a) Calculate the value of the equilibrium constant for the reaction at 298 K. (b) Calculate the standard entropy change, ASo, for the reaction at 298 K. (c) If CIF, was produced as a liquid rather than a gas, how would the sign and magnitude AS for the reaction be affected? Explain. (d) At 298 K the absolute entropies of Cl:(8) and CIF, (8) are 223 J/molK and 282 J/molK., (e) Calculate the value of the absolute entropy of Fe (g) at 298 K. respectively. Account for the larger entropy value of CIFs (8) relative to that of C2 (8).Explanation / Answer
Answer:
a) /Gorxn= n /Gof products - n /Gof reactants
/Gorxn= 2 mol ( -123.0 kJ mol -1 ) - 1 mol ( 0 kJ mol -1 ) - 3 mol (0 kJ mol -1 )
/Gorxn= - 246.0 kJ
/Gorxn= - 2.303 RT log K
K = antilog ( /Gorxn / - 2.303 RT)
- 246.0 kJ
K = antilog (--------------------------------------------------------) = 1.30 x 10 43 (AnS)
- 2.303 x 0.008314 kJ mol -1 K -1 x 298 K
b)/Horxn= 2 mol /Hfo ClF3 - 1 mol /Hfo Cl2 - 3 mol/HfoF2/Horxn= n /Hof products - n /Hof reactants
/Horxn= 2 mol ( -163.2 kJ mol -1 ) - 1 mol ( 0 kJ mol -1 ) - 3 mol (0 kJ mol -1 )
/Horxn= - 326.4 kJ
/Gorxn= /Horxn - T/Sorxn Solve for /Sorxn
/Horxn -/Gorxn - 326.4 kJ - ( - 246.0 kJ)
/Sorxn = ------------------------- = ------------------------------------- = - 0.270 kJ / K = - 270. J / K(Ans)
T 298 K
C) / = nproduct gases - nreactant gases
Cl2(g) + 3F2(g) ---> 2ClF3(g) /n = 2 mol - 4 mol = -2 mol
Cl2(g) + 3F2(g) ---> 2ClF3(l) /n = 0 mol - 4 mol = -4 mol
The sign for /So would remain negative because the sign of /n remains the same. However, since the value for /n doubles, the size of /So should also double.
d)The Cl2 molecule is a symmetrical linear molecule while ClF3 is a nonsymmetrical bent T molecule. The motions allowed ClF3 have more freedom of movement increasing the entropy.
e) /Sorxn = 2 mol SoClF3 - 1 mol SoCl2 - 3 mol SoF2 /Sorxn= n Sof products - n Sof reactants
2 mol SoClF3 - 1 mol SoCl2 - /Sorxn
SoF2 = ------------------------------------------------------
3 mol
2 mol x 281.50 J/mol K - 1 mol x 222.96 J/mol K - ( - 270. J / K )
SoF2 = ----------------------------------------------------------------------------------------
3 mol
= 203 J/mol K
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