Following is the - complete Answer -&- Explanation: for the given: Question: in...typed format...
Answer:
Horxn. = - 366.0 kJ /
mol
Horxn. = - 200.09 kJ /
mol
Horxn. = - 132.5 kJ /
mol
Horxn. = - 114.18 kJ /
mol
Explanation:
Following is the complete : Explanation: for the above: Answer...
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Part
- (a):
CH3CHO (g) + O2 (g)
-------------Equation - 1Now, we know the following standard enthalpy of formation: of the given: reactants and products:
Hfo
[ C2H4 (g) ] = + 52.3 kJ /
mol
Hfo
[ O3 (g) ] = + 143.0 kJ / mol
Hfo
[ CH3CHO(g) ] = -170.7 kJ /
mol
Hfo
[O2(g) ] = 0.0 kJ / molTherefore, according to Equation - 1: the value of the standard change of enthalpy: of Equation - 1, will be the following;

Horxn.
= [ ( 1.0 mol ) x (
Hfo [ CH3CHO(g))
] ) + (1.0 mol ) x (
Hfo
[O2(g) ] ) ] - [ (
Hfo[
C2H4 (g) ] +
Hfo[ O3 (g) )
]
Horxn. = [ ( 1.0 mol ) x
(-170.7 kJ / mol ) ] ) + (1.0 mol ) x (0.0 kJ/mol )
] - [ ( 1.0 mol) x (52.3 kJ / mol
] + (1.0 mol ) x ( 143.0 kJ / mol ) ] = - 366.0 kJ /
mol
Horxn. = - 366.0 kJ /
mol
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Part
- (b):
NO2 (g) + O2 (g)
-------------Equation - 2Now, we know the following standard enthalpy of formation: of the given: reactants and products:
Hfo
[ NO2 (g) ] = + 33.2 kJ / mol
Hfo
[ O3 (g) ] = + 143.0 kJ / mol
Hfo
[NO(g) ] = +90.29 kJ / mol
Hfo
[O2(g) ] = 0.0 kJ / molTherefore, according to Equation - 1: the value of the standard change of enthalpy: of Equation - 1, will be the following;

Horxn.
= [ ( 1.0 mol ) x (
Hfo [ NO2(g)) ] )
+ (1.0 mol ) x (
Hfo
[O2(g) ] ) ] - [ (
Hfo[ O3 (g) ] +
Hfo[ NO(g) )
]
Horxn. = [ ( 1.0 mol ) x (+
33.2 kJ / mol ) ] ) + (1.0 mol ) x (0.0 kJ/mol ) ]
- [ ( 1.0 mol) x (+ 143.0 kJ / mol ] +
(1.0 mol ) x ( +90.29 kJ / mol ) ] = - 200.09 kJ /
mol
Horxn. = - 200.09 kJ /
mol
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Part
- (c):
H2SO4 (aq) -------------Equation -
3Now, we know the following standard enthalpy of formation: of the given: reactants and products:
Hfo
[H2SO4 (aq) ] = -814.0 kJ /
mol
Hfo
[ SO3 (g) ] = −395.7 kJ /
mol
Hfo
[H2O (l) ] = −285.8 kJ /
molTherefore, according to Equation - 1: the value of the standard change of enthalpy: of Equation - 1, will be the following;

Horxn.
= [ ( 1.0 mol ) x (
Hfo
[H2SO4 (aq) ] ) ] -
[ (
Hfo[ SO3 (g) ] +
Hfo[ H2O (l) )
]
Horxn. = [ ( 1.0 mol ) x
(-814.0 kJ / mol ) ] ) ] - [ (
1.0 mol) x (−395.7 kJ / mol ) + (1.0 mol ) x (−285.8 kJ /
mol ) ]
= - 132.5 kJ / mol
Horxn. = - 132.5 kJ /
mol
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Part
- (d):
2 NO2 (g) -------------Equation -
3Now, we know the following standard enthalpy of formation: of the given: reactants and products:
Hfo
[ NO2 (g) ] = + 33.2 kJ / mol
Hfo
[NO(g) ] = +90.29 kJ / mol
Hfo
[O2(g) ] = 0.0 kJ / molTherefore, according to Equation - 1: the value of the standard change of enthalpy: of Equation - 1, will be the following;

Horxn.
= [ ( 2.0 mol ) x (
Hfo [NO2 (g) ] )
] - [ ( 2.0 mol ) x (
Hfo[ NO(g) ] + ( 1.0 mol ) x
(
Hfo[
O2 (g)) ]
Horxn. = [ ( 2.0 mol ) x (+
33.2 kJ / mol ) ] ) ] - [ (2.0
mol) x (+90.29 kJ / mol ) + (1.0 mol ) x (0.0 kJ/mol)
]
= - 114.18 kJ / mol
Horxn. = - 114.18 kJ /
mol
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Q1
Q2
Q3
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