At what temperature will the following be spontaneous? Delta H = 18 kj and delta
ID: 506346 • Letter: A
Question
At what temperature will the following be spontaneous? Delta H = 18 kj and delta S = -60 J/K delta H = + 18 kJ and delta S + 60 J/K delta H +18 kJ and delta S = 60 J/K Predict the sign of delta S for the reaction HCI (g) H + (aq) + Cl - (aq) Using appendix 4, calculate delta S, delta H and delta G for the following reaction: CH_4 (g) + 20_2 (g) CO_2 (g) + 2H_2O (g) For the reaction: 2NO_2 (g) N_2O_4 (g) The value of delta H = -58.03 kJ and delta S = 176 J/K Calculate delta G at 298K Assuming AH and AS are unaffected by temperature, at what temperature is delta G = 0? Calculate AG for an acid ionization if Ka = 4.5 times 10^-3 at 25 degree CExplanation / Answer
Question 1.
Apply Gibbs Free Energy concept
in which
dG < 0, will favour a forward/spotaneous process
case
a)
dH = -18 kJ/mol = -18000 J/mol
dS = -60 J/K
dG = dH - T*dS
dG < 0 so
dG - T*dS < 0
-18000 - T*(-60) < 0
-18000 < -60T
18000/60 > T
T < 300 K, the temperature must be low in order to favour this (entropy is negative)
b)
similar
dH = 18 kJ/mol = +18000 J/mol
dS = +60 J/K
dG = dH - T*dS
dG < 0 so
dG - T*dS < 0
18000 - T*(60) < 0
18000 < 60T
18000/60 < T
T > 300 K, the temperature must be high in order to favour this (entropy is positive)
c)
finally
dH = 18 kJ/mol = 18000 J/mol
dS = -60 J/K
dG = dH - T*dS
dG < 0 so
dG - T*dS < 0
18000 - T*(-60) < 0
18000 < -60T
-18000/60 > T
T < -300 K, this will never be favoured, since T can't be negative, so this process is impossible
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