Standard molar enthalpy of formation of `CO_(2)` is equal to `:`

Options

A. 5
B. 10
C. 95
D. 100

Answer

B

Explanation

`A hArr B`
`DeltaG^@ = DeltaH^@ - T DeltaS^@ `,
`DeltaG^@ = 0-2.303 RT log_10 K`
` -2.303 RT log_10 K = DeltaH^@ - T DeltaS^@`,
`2.303 RT log_10 K = T DeltaS^@ - DeltaH^@`
`log_10 K = (T DeltaS^@ - DeltaH^@)/(2.303 RT) = (298 xx 10 + 54.07 xx 1000)/(2.303 xx 8.314 xx 298) = 10`.