When the oxide of generic metal M is heated at 25.0 °C, only a negligible amount

Question

When the oxide of generic metal M is heated at 25.0 °C, only a negligible amount of M is produced.

Question 22 of 31 Ma Sapling Learning macmillan learning When the oxide of generic metal M is heated at 25.0 °C, only a negligible amount of M is produced M02(s) M(s) +02(g) Go=287.6 kJ/mol When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. What is the chemical equation of this coupled process? Show that the reaction is in equilibrium, include physical states, and represent graphite as C(s) What is the thermodynamic equilbrium constant for the coupled reaction? Number K=

Explanation / Answer

the reaction given MO2(s)<----> M(s)+ O2(g), deltaGo= 287.6 Kj/mole   (1)

since deltag is +ve, the reaction is not spontaneous.

this is combines with C(S)+ O2 (g)------->CO2(g), deltaGo= -393.5 Kj/mole            (2)

combining the reaction- 1 and 2 gives MO2(s)+ C (s)--------> M (s)+CO2(g), deltaGo= 287.6-393.5=-105.9 j/mole

since deltaG is -ve, the reaction is spontaneous

deltaGo= -RTlnK

=-105.9*1000 J/mole= -8.314 (j/mole.*298*lnK

lnK= 105.9*1000/(8.314*298)=42.74, K= 3.658*1018

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