For the reaction: 4Ag(s, pure) + O_{2}(g) βΆ 2Ag_{2}O(s, pure), the standard enthalpy change, βπ»^{0} = β 61080 J, and the standard entropy change, βπ^{0} = β 132.22 J K^{β1}, in the temperature range from 298 K to 500 K.

The temperature above which Ag_{2}O 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