-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
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.
To find the standard enthalpy change of formation of butanone, we need to use the given equation for the complete combustion of butanone and the standard enthalpies of formation for carbon dioxide (CO₂) and water (H₂O).
The equation for the combustion of butanone is:
C₄H₈O + 6O₂ → 4CO₂ + 4H₂O
The change in enthalpy for the given equation is -2440 kJ/mol, which is the enthalpy of combustion. This means that 2440 kJ/mol of energy is released when one mole of butanone is completely burned.
From the equation, we can see that 4 moles of CO₂ and 4 moles of H₂O are formed for every mole of butanone burned.
The standard enthalpies of formation for CO₂ and H₂O are given as -394 kJ/mol and -286 kJ/mol, respectively.
To calculate the standard enthalpy change of formation for butanone, we need to use the following equation:
ΔH°f = ΣnΔH°f(products) - ΣmΔH°f(reactants)
Where ΔH°f is the standard enthalpy change of formation, and n and m are the stoichiometric coefficients.
For butanone, there is only one mole of butanone in the reaction, so we have:
ΔH°f = (4×ΔH°f(CO₂)) + (4×ΔH°f(H₂O)) - ΔH°c
Plugging in the values:
ΔH°f = (4×-394 kJ/mol) + (4×-286 kJ/mol) - (-2440 kJ/mol)
ΔH°f = -1576 kJ/mol + -1144 kJ/mol + 2440 kJ/mol
ΔH°f = -280 kJ/mol
Therefore, the standard enthalpy change of formation of butanone is -280 kJ/mol.
So, the correct answer is option A) -280.
To determine the standard enthalpy change of formation of butanone, we need to use Hess's law. According to Hess's law, the change in enthalpy for a reaction can be calculated by manipulating other reactions with known enthalpy changes.
First, let's analyze the given equation:
CH3COCH2(g) + 5½O2(g) → 4CO2(g) + 4H2O(1) ΔH = -2440 kJ mol⁻¹
This equation represents the complete combustion of butanone (CH3COCH2). We need to find the standard enthalpy change of formation of butanone (ΔHf).
The standard enthalpy change of formation (ΔHf) is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
To find the standard enthalpy change of formation of butanone, we need to manipulate the given equation and other known enthalpy changes to cancel out the CO2 and H2O terms, leaving only the butanone.
We can use the following known reactions:
C(graphite) + O2(g) → CO2(g) ΔH = -394 kJ mol⁻¹ (Given)
H2(g) + ½O2(g) → H2O(1) ΔH = -286 kJ mol⁻¹ (Given)
To cancel out the CO2 term, we multiply the first equation by 4:
4C(graphite) + 4O2(g) → 4CO2(g) ΔH = -4 × (-394) kJ mol⁻¹ = +1576 kJ mol⁻¹
To cancel out the H2O term, we multiply the second equation by 4:
4H2(g) + 2O2(g) → 4H2O(1) ΔH = 4 × (-286) kJ mol⁻¹ = -1144 kJ mol⁻¹
Since we want to end up with butanone and the coefficients in the given equation are already balanced, we don't need to manipulate it any further.
Now, we can add the manipulated reactions and the given equation to get the overall reaction that allows us to calculate the enthalpy change of formation of butanone:
4C(graphite) + 4H2(g) + CH3COCH2(g) + 4½O2(g) → 4CO2(g) + 4H2O(1) ΔH = 1576 + (-1144) + (-2440) kJ mol⁻¹
Simplifying:
CH3COCH2(g) + 5½O2(g) → 4CO2(g) + 4H2O(1) ΔH = -1008 kJ mol⁻¹
The standard enthalpy change of formation of butanone (ΔHf) is the enthalpy change for the formation of one mole of butanone from its elements in their standard states. Since the stoichiometric coefficient of butanone is 1 in the balanced equation, the enthalpy change can be directly obtained from the overall reaction:
ΔHf = -1008 kJ mol⁻¹
Therefore, the correct option is:
Α) -280