Question

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

Answer 1

Answer:

The value of $\Delta H$ for the desired reaction will be 249.75 KJ.

Explanation:

The desired reaction is shown below

$\textrm{PCl}_{5}\left ( g \right )\rightarrow \textrm{PCl}_{3}\left ( g \right )+\textrm{Cl}_{2}\left ( g \right )$

The desired reaction can be obtained by adding the given reactions and then dividing both sides by 4.

$P_{4}\left ( s \right )+6Cl_{2}\left ( g \right )\rightarrow 4PCl_{3}\left ( g \right ) \\4PCl_{5}\left ( g \right )\rightarrow P_{4}\left ( s \right )+10Cl_{2}\left ( g \right )$

Net Enthalpy change for the desired reaction

$\displaystyle \frac{3438-2439}{4} \textrm{ KJ} = 249.75 \textrm{ KJ}$

$\Delta H = 249.75 \textrm{ KJ}$

Answer 2

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:

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