For the given system, (a) with dG = 0 dG = dH - TdS dH = TdS T (temperature) = d

Question

For the given system,

(a) with dG = 0

dG = dH - TdS

dH = TdS

T (temperature) = dH/dS

with,

dH = -14.9 kJ/mol

dS = -50 J/K.mol = -0.05 kJ/K.mol

we get,

Temperature (T) = -14.9/-0.05 = 298 K

So the reaction would be spontaneous below 298 K

(b) dG = dH - TdS

Using Van't Hoff relation,

lnK = -dH/RT + dS/R

with K = 3.0

R = gas constant

feeding dH and dS from above,

ln(3.0) = -(-14900/8.314 x T) + (-50/8.314)

ln(3.0) = = 1792.16/T - 6.014

Temperature (T) = 252 K

                            = 252 - 273 = -21 oC

   

Explanation / Answer

t 8 Page 7 of 8 (8 points) For a certain chemical reactionAH--14.9 kJ/mol. AS-50 JK-mol, (a) what is the rature (in °C) range for the reaction to be spontaneous? (b) Assume that the values of AH and AS are dependent of temperature. If the equilibrium constant, K -3.0, what is the temperature in C

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