To find the standard enthalpy change of formation of butanone, we need to use the equation:
4CO₂(g) + 4H₂O(1) → CH₃COCH₃(1)
The standard enthalpy change of formation of a substance is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
According to the given data, the enthalpy change of formation of CO₂ is -394 kJ/mol and the enthalpy change of formation of H₂O is -286 kJ/mol.
Using the equation for the complete combustion of butanone, we can see that 4 moles of CO₂ and 4 moles of H₂O are formed for every mole of butanone. Therefore, the enthalpy change of formation of butanone can be calculated using the enthalpy changes of formation of CO₂ and H₂O.
Enthalpy change of formation of butanone = 4 * (-394 kJ/mol) + 4 * (-286 kJ/mol) = -1576 kJ/mol - 1144 kJ/mol = -2720 kJ/mol
Therefore, the standard enthalpy change of formation of butanone is approximately -2720 kJ/mol.
None of the given options (-280, +280, -1760, +1760) are equal to this result.
-The equation for the complete combustion of butanone, C,H,COCH,, is
CHCOCH₂(1) + 5½0₂(g) → 4CO₂(g) + 4H₂O(1) AH = -2440 kJ mol-¹
Substance:-
CO₂(g)
H₂O(1)
AH, /kJ mol-1:-
-394
-286
From the above data, the standard enthalpy change of formation of butanone, in kJ mol-¹, is
Α) -280
B )+280
с )-1760
d) +1760
1 answer