Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:





P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°rxn = ?





Given:





PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn= +157 kJ





P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ





A) -1835 kJ


B) -1364 kJ


C) -1050. kJ


D) -1786 kJ


E) -2100. kJ

How would you manipulate the two reactions to get to the desired reaction?
In the reactants of the wanted reaction, you have 1 P4. The second reaction has 1 P4 in the reactants, so you don't want to change that reaction in any way.
In the products of the wanted reaction, you have 4PCL5. In the first reaction, you have 1 PCL5 in the reactants side. So, you need to multiply it by four, and also change it so that it's on the products side (multiply by -1). Everything else also gets into it's right place after you do this.

In conclusion,
multiply the enthalpy of the first reaction by -4, and don't change the second one.

(+157)*-4 + (-1207)
= -1835

Answer is A

P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°rxn = ?

Is a combination of the two reactions. Some are flipped:

Cl2 + PCl3 => PCl5 -157 kJ
P4 + 6Cl2 => 4PCl3 -1207 kJ

Now we need 10 Cl2 total and need to cancel out the PCl3 with 4. So multiply the reaction by 4.
4Cl2 + 4PCl3 => 4PCl5 = -628 kJ
6 Cl2 + P4 => 4PCl3 = -1207 kJ\
The PCl3 cancels when you add the reaction and the 4 an 6 Cl2 become 10 Cl2.
Final answer: A) -1835 kJ