Answer:
a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ, ΔSºrxn = 0.113 kJ/K
b) At 753.55 ºC or higher
c )ΔG = 1.8 x 10⁴ J
K = 8.2 x 10⁻²
Explanation:
a) C6H5−CH2CH3 ⇒ C6H5−CH=CH2 + H₂
ΔHf kJ/mol -12.5 103.8 0
ΔGºf kJ/K 119.7 202.5 0
Sº J/mol 255 238 130.6*
Note: This value was not given in our question, but is necessary and can be found in standard handbooks.
Using Hess law to calculate ΔHºrxn we have
ΔHºrxn = ΔHfº C6H5−CH=CH2 + ΔHfº H₂ - ΔHºfC6H5−CH2CH3
ΔHºrxn = 103.8 kJ + 0 kJ - (-12.5 kJ)
ΔHºrxn = 116.3 kJ
Similarly,
ΔGrxn = ΔGºf C6H5−CH=CH2 + ΔGºfH₂ - ΔGºfC6H5CH2CH3
ΔGºrxn= 202.5 kJ + 0 kJ - 119.7 kJ = 82.8 kJ
ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K
b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using
ΔGrxn = ΔHrxn -TΔS
we see that will happen when the term TΔS becomes greater than ΔHrxn since ΔS is positive , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that ΔºHrxn and ΔSºrxn remain constant at the higher temperature and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.
0 = 116 kJ -T (0.113 kJ/K)
T = 1026.5 K = (1026.55 - 273 ) ºC = 753.55 ºC
c) Again we will use
ΔGrxn = ΔHrxn -TΔS
to calculate ΔGrxn with the assumption that ΔHº and ΔSºremain constant.
ΔG = 116.3 kJ - (600+273 K) x 0.113 kJ/K = 116.3 kJ - 873 K x 0.113 kJ/K
ΔG = 116.3 kJ - 98.6 kJ = 17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )
Now for solving for K, the equation to use is
ΔG = -RTlnK and solve for K
- ΔG / RT = lnK ∴ K = exp (- ΔG / RT)
K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K x 873 K)) = 8.2 x 10⁻²
= 0.0319 mol AgNO₃