Given the following data:
4PCl5(s) �¨ P4(s) + 10Cl2(g) ƒ¢H = +1774 kJ
P4(s) + 6Cl2(g) �¨ 4PCl3(l) ƒ¢H = -1278.8 kJ
Find the ĢH of the following reaction:
PCl5(s) �¨ PCl3(l) + Cl2(g)
To find the enthalpy change (ĢH) of the given reaction:
Step 1: Recognize that the given reaction is not given directly, but it can be achieved by subtracting the second given reaction from the first reaction. Therefore, we need to reverse the second reaction.
P4(s) + 10Cl2(g) �¨ 4PCl5(s) ƒ¢H = -1774 kJ (Multiplying the reaction by -1 to reverse it)
Now the two reactions can be added together:
4PCl5(s) �¨ P4(s) + 10Cl2(g) ƒ¢H = +1774 kJ
P4(s) + 10Cl2(g) �¨ 4PCl5(s) ƒ¢H = -1774 kJ
(summing these two reactions)
This will cancel out the 4PCl5(s) and 4PCl5(s), leaving us with:
P4(s) + 10Cl2(g) �¨ P4(s) + 10Cl2(g) ƒ¢H = +1774 kJ - 1774 kJ
Step 2: Simplifying the equation, we get:
P4(s) + 10Cl2(g) �¨ 0 kJ
Therefore, the ĢH of the reaction:
PCl5(s) �¨ PCl3(l) + Cl2(g)
is 0 kJ.