For the reaction: 4Ag(s, pure) + O2(g) βΆ 2Ag2O(s, pure), the standard enthalpy change, βπ»0 = β 61080 J, and the standard entropy change, βπ0 = β 132.22 J Kβ1, in the temperature range from 298 K to 500 K.
The temperature above which Ag2O decomposes in an atmosphere containing oxygen at a partial pressure ππ2Β Β Β = 0.3 atm is Β Β Β Β Β Β Β Β Β Β Β Β (in K to one decimal place).
Given: Gas constant R = 8.314 J K-1 mol-1
Delta G0 = Delta H0 - T delta S0 Also Delta G = Delta G0 + RTlnk In case of delta G =0 Delta G0= -RTlnkeq -RTlnkeq = delta H0 -T delta S0 -8.314T ln(1/0.3) = -61080 -T (-132.22) T = 499.75 K
Delta G0 = Delta H0 – T delta S0
See lessAlso Delta G = Delta G0 + RTlnk
In case of delta G =0
Delta G0= -RTlnkeq
-RTlnkeq = delta H0 -T delta S0
-8.314T ln(1/0.3) = -61080 -T (-132.22)
T = 499.75 K


See lessCorrect answer is 429.5 K. @Digbijaya igit described it best. Thank you.
Correct answer is 429.5 K. digbijaya igit igit described it best. Thank you.
Delta G0 = Delta H0 β T delta S0 Also Delta G = Delta G0 + RTlnk In case of delta G =0 Delta G0= -RTlnkeq -RTlnkeq = delta H0 -T delta S0 -8.314T ln(1/0.3) = -61080 -T (-132.22) T = 499.75 K
Delta G0 = Delta H0 β T delta S0
See lessAlso Delta G = Delta G0 + RTlnk
In case of delta G =0
Delta G0= -RTlnkeq
-RTlnkeq = delta H0 -T delta S0
-8.314T ln(1/0.3) = -61080 -T (-132.22)
T = 499.75 K