Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: PCl5(g) → PCl3(g) + Cl2(g) P4(s) + 6Cl2(g) → 4PCl3(g) ΔH = -2439 kJ
4PCl5(g) → P4(s) + 10Cl 2(g) ΔH = 3438 kJ
Answers
The value of ΔH will be 249.75 KJ.
To obtain the reaction :
PCl5(g) → PCl3(g) + Cl2(g)
We have to add the given reactions and then divide it by 4.
(i) adding the reactions
P4(s) + 6Cl2(g) → 4PCl3(g) ΔH = -2439 kJ
4PCl5(g) → P4(s) + 10Cl 2(g) ΔH = 3438 kJ
we get:
4PCl5(g) → 4PCl3(g) + 4Cl2(g), ΔH = 3438 - 2439
(ii) dviding by 4
PCl5(g) → PCl3(g) + Cl2(g) , ΔH = (3438 - 2439)/4
ΔH = 249.75 kJ is the required enthapy.
Learn more about enthalpy:
Answer:
The value of for the desired reaction will be 249.75 KJ.
Explanation:
The desired reaction is shown below
The desired reaction can be obtained by adding the given reactions and then dividing both sides by 4.
Net Enthalpy change for the desired reaction
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