Use the standard reaction enthalpies given below in steps 1 and 2 to determine ΔHorxn for the following reaction: P4(g) + 10 Cl2(g) ---> 4 PCl5(s) ΔHorxn = ________ kJ Given: (1) PCl5(s) --> PCl3(g) + Cl2(g) ΔHorxn = +157 kJ (2) P4(g) + 6 Cl2(g) --> 4 PCl3(g) ΔHorxn = -1207 kJ
Question
Use the standard reaction enthalpies given below in steps 1 and 2 to determine ΔHorxn for the following reaction: P4(g) + 10 Cl2(g) ---> 4 PCl5(s) ΔHorxn = ________ kJ Given: (1) PCl5(s) --> PCl3(g) + Cl2(g) ΔHorxn = +157 kJ (2) P4(g) + 6 Cl2(g) --> 4 PCl3(g) ΔHorxn = -1207 kJ
Solution
To find the enthalpy change (ΔH) for the reaction P4(g) + 10 Cl2(g) --> 4 PCl5(s), we can use the given reactions and their enthalpies.
First, we need to manipulate the given reactions to match the desired reaction.
The first given reaction is PCl5(s) --> PCl3(g) + Cl2(g) with ΔH = +157 kJ. We need 4 PCl5(s) for our desired reaction, so we multiply the entire reaction by 4:
4 PCl5(s) --> 4 PCl3(g) + 4 Cl2(g) with ΔH = +157 kJ * 4 = +628 kJ.
The second given reaction is P4(g) + 6 Cl2(g) --> 4 PCl3(g) with ΔH = -1207 kJ. This reaction can be used as it is.
Now, we add these two manipulated reactions:
4 PCl5(s) --> 4 PCl3(g) + 4 Cl2(g) (+628 kJ) P4(g) + 6 Cl2(g) --> 4 PCl3(g) (-1207 kJ)
P4(g) + 10 Cl2(g) --> 4 PCl5(s)
The enthalpy change of the desired reaction is the sum of the enthalpy changes of the manipulated reactions:
ΔH = +628 kJ - 1207 kJ = -579 kJ.
So, the enthalpy change for the reaction P4(g) + 10 Cl2(g) --> 4 PCl5(s) is -579 kJ.
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