The standard enthalpy of formation of `CO (g), CO_2 (g) , N_2O (g) " and " N_2O_4 (g)` are respectively - 10 . 393, 81 and -10 kJ `"mol"^(-1)` . Enthalpy change (in kJ) of the follwing reaction is
`N_2O_4(g) + 3CO (g) to N_2O(g) + 3CO_2 (g)`

Options

A. `-1058`
B. `+1058`
C. `-957`
D. `+957`

Answer

A

Explanation

Given , standard enthalpy of formation of `CO = -10 ` kJ/mol
Standard enthalpy of formation of `CO_2` = -393 kJ/mol
Standard enthalpy of formation of `N_2O` = 81 kJ/mol
Standard enthalpy of formation of `N_2O_4` = - 10 kJ/mol
`N_2O_4 + 3CO to N_2O + 3CO_2`
`Delta H` (enthalpy change )` = sum Delta H_(N_2O) + 3Delta H_(CO_2) - sum Delta H_(N_2O_4) + 3Delta H_(CO)`
` = [sum 81 + (3 xx - 393 ) - sum (-10 + (3 xx -10) ]`
` Delta H = - 1058` kJ/mol