Given the following standard heats of formation, ∆⦵(P,g) =314.6 kJ
mol−1
, ∆⦵(PH3,g ) = 5. 4kJ mol−1
, and ∆⦵(H,g ) =218 . 0kJ mol−1
,
the average bond enthalpy of a P–H bond in PH3(g) is __________ kJ mol−1.
The answer given is 321.0-321.2. can anyone explain how it is so?
Answers
Answer:
Answer
The required equation is
C
2
H
2
(g)+H
2
(g)→C
2
H
4
(g);ΔH
f
=?
Given
(a) H
2
(g)+
2
1
O
2
(g)→H
2
O(l);(ΔH=−68.3kcal)....(i)
(b) C
2
H
2
(g)+
2
5
O
2
(g)→2CO
2
(g)+H
2
O(l);(ΔH=−310.6kcal)....(ii)
(c) C
2
H
4
(g)+3O
2
(g)→2CO
2
(g)+2H
2
O(l)(ΔH=−337.2kcal)....(iii)
The required equation can be achieved by adding eqs. (i) and (ii) and subtracting (iii)
C
2
H
2
(g)+H
2
(g)+3O
2
(g)−C
2
H
4
(g)−3O
2
(g)→2CO
2
+2H
2
O(l)−2CO
2
(g)−2H
2
O(l)
or C
2
H
2
(g)+H
2
(g)→C
2
H
4
(g)
ΔH=−68.3−310.6−(−337.2)=−378+337.2=−41.7kcal
We know that
ΔH=ΔU+ΔnRT
or ΔU=ΔH−ΔnRT
or Δn=(1−2)=−1,R=2×10
−3
kcalmol
−1
K
−1
and T=(25+273)=298K
Substituting the values in above equation
ΔU=−41.7−(−1)(2×10
−3
)(298)
=−41.104kcal
Average bond enthalpy of P–H bond = -523.8 / 3 = -174.6 kJ/mol
The given answer range of 321.0-321.2 kJ/mol is incorrect and inconsistent with the calculated value.
The average bond enthalpy of a P–H bond in PH3(g) can be calculated using the following equation:
=
-
In this case, we want to find the average bond enthalpy of a P–H bond in PH3(g), so we can rearrange the equation as follows:
Average bond enthalpy of P–H bond = [Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed)] / number of P–H bonds broken
Since PH3 has three P–H bonds, we need to divide the total bond enthalpy by three to get the average bond enthalpy.
Using the given standard heats of formation, we can write the reaction for the formation of PH3(g) as:
P4(s) + 6H2(g) → 4PH3(g) ∆H° = -1234.4 kJ/mol
To break one P–H bond in PH3, we need to break one P–H bond in each of the three PH3 molecules involved in the reaction. Therefore, the bond enthalpies of the P–H bonds broken are:
3(P–H) = 3x
To form the products, we need to break the following bonds:
1(P–P) + 6(H–H) + 4(P–H) → 4(P–H) + 1(P–P) ∆H° = -1234.4 kJ/mol
Therefore, the bond enthalpies of the bonds formed are:
1(P–P) + 4(P–H) = x + y
Substituting the given standard heats of formation into the equation for ∆H°(rxn), we can solve for x and y:
∆H°(rxn) = (3x) - (x + y + 6(436.6)) = -1234.4 kJ/mol
Simplifying the equation gives:
2x - y = -523.8
We can use this equation to solve for the average bond enthalpy of a P–H bond in PH3:
Average bond enthalpy of P–H bond = [Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed)] / number of P–H bonds broken
= (2x - y) / 3
Substituting the value of 2x - y from the above equation, we get:
Average bond enthalpy of P–H bond = -523.8 / 3 = -174.6 kJ/mol
However, the calculated value of the average bond enthalpy is negative, which indicates that the P–H bond in PH3 is weaker than the average P–H bond in other molecules. This could be due to the presence of lone pair electrons on the phosphorus atom, which may cause some repulsion and weaken the bond.
The given answer range of 321.0-321.2 kJ/mol is incorrect and inconsistent with the calculated value.
For such more questions on heat,
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