You wish to know the enthalpy change for the formation of liquid PCl3 from the elements shown below.

P4(s) + 6 Cl2(g)--> 4 PCl3(l) ΔH°f = ?
The enthalpy change for the formation of PCl5 from the elements can be determined experimentally, as can the enthalpy change for the reaction of PCl3(l) with more chlorine to give PCl5(s).

P4(s) + 10 Cl2(g)--> 4 PCl5(s) ΔH°rxn= -1774.0 kJ

PCl3(l) + Cl2(g)--> PCl5(s) ΔH°rxn = -123.8 kJ
Use these data to calculate the enthalpy change for the formation of 1.50 mol of PCl3(l) from phosphorus and chlorine.

P4(s) + 10 Cl2(g)--> 4 PCl5(s) ΔH°rxn= -1774.0 kJ

PCl3(l) + Cl2(g)--> PCl5(s) ΔH°rxn = -123.8 kJ

Use eqn 1 as is. Multiply eqn 2 by 4 (multiply delta H rxn by 4 also), reverse it (change the sign of delta H x 4), and add to eqn 1. That will give you the reaction you want. Add the delta Hs to arrive at delta H for the desired reaction. That will be delta H for 4 moles of PCl3 and you change that to 1.50 moles PCl3.

Well, it seems we have a PCl3 party going on here! Let's calculate the enthalpy change for the formation of 1.50 mol of PCl3(l) from phosphorus and chlorine.

First, we need to find the enthalpy change for the reaction given:

P4(s) + 6 Cl2(g) --> 4 PCl3(l) ΔH°f = ?

Since we don't have the direct enthalpy change for this reaction, we'll take a detour. We know the enthalpy change for the formation of PCl5 from the elements is -1774.0 kJ, and the enthalpy change for the reaction of PCl3(l) with more chlorine to give PCl5(s) is -123.8 kJ.

Now, let's do some math tricks! Since we want to find the enthalpy change for 1.50 mol of PCl3(l), we can consider the reaction between PCl3(l) and Cl2(g):

PCl3(l) + Cl2(g) --> PCl5(s) ΔH°rxn = -123.8 kJ

Now, if we double this reaction to get 2 moles of PCl3(l), the enthalpy change will also be doubled:

2 PCl3(l) + 2 Cl2(g) --> 2 PCl5(s) ΔH°rxn = -247.6 kJ

However, we want 1.50 mol of PCl3(l), not 2 moles. So we divide the enthalpy change by 2 and multiply by 1.5 to get the final value:

(1.50 mol / 2 mol) * (-247.6 kJ / 2) = -185.7 kJ

So, the enthalpy change for the formation of 1.50 mol of PCl3(l) from phosphorus and chlorine is approximately -185.7 kJ. Keep in mind, these calculations are subject to accuracy contaminated by my delightful clown humor!

To calculate the enthalpy change for the formation of 1.50 mol of PCl3(l) from phosphorus and chlorine, we need to use the given enthalpy changes and stoichiometric ratios.

First, let's start by looking at the reaction between P4(s) and 6 Cl2(g) to form 4 PCl3(l). The enthalpy change for this reaction is not given directly, but we can use the enthalpy change for the formation of PCl5 from the elements, as well as the enthalpy change for the reaction of PCl3(l) with Cl2(g) to form PCl5(s), to calculate it.

The enthalpy change for the formation of PCl5 from the elements is -1774.0 kJ, and the enthalpy change for the reaction of PCl3(l) with Cl2(g) to form PCl5(s) is -123.8 kJ.

Since the reaction of forming 4 mol of PCl3(l) from P4(s) and 6 Cl2(g) gives 1 mol of PCl5(s), we can write the following equation using the given enthalpy changes:

4 PCl3(l) + Cl2(g) --> PCl5(s) ΔH°rxn = -123.8 kJ

From this equation, we can see that the enthalpy change for the formation of 4 mol of PCl3(l) is -123.8 kJ.

Next, we need to calculate the enthalpy change for the formation of 1 mol of PCl3(l). To do this, we divide the enthalpy change by 4:

ΔH°formation of 1 mol of PCl3(l) = (-123.8 kJ) / 4 = -30.95 kJ/mol

Finally, we can calculate the enthalpy change for the formation of 1.50 mol of PCl3(l) by multiplying the enthalpy change for the formation of 1 mol by the number of moles:

ΔH°formation of 1.50 mol of PCl3(l) = (-30.95 kJ/mol) * 1.50 mol = -46.43 kJ

Therefore, the enthalpy change for the formation of 1.50 mol of PCl3(l) from phosphorus and chlorine is -46.43 kJ.

To calculate the enthalpy change for the formation of 1.50 mol of PCl3(l) from phosphorus and chlorine, we can use the given enthalpy changes for the reaction of PCl3 with chlorine and the reaction of PCl5 formation.

Let's break down the process step by step:

Step 1: Calculate the enthalpy change for the reaction of PCl3 with chlorine to give PCl5

We are given: PCl3(l) + Cl2(g) --> PCl5(s) ΔH°rxn = -123.8 kJ

To calculate the enthalpy change for an equation, we need to multiply the reaction by a coefficient such that the number of moles of the desired product matches the stoichiometry of the target reaction. In this case, we want to find the enthalpy change for the formation of 4 moles of PCl5.

So, we multiply the given reaction by 4:
4 (PCl3(l) + Cl2(g) --> PCl5(s)) ΔH°rxn = -123.8 kJ x 4

Step 2: Calculate the enthalpy change for the reaction of PCl5 formation

We are given: P4(s) + 10 Cl2(g) --> 4 PCl5(s) ΔH°rxn = -1774.0 kJ

To calculate the enthalpy change for the formation of 4 moles of PCl5, we divide the given enthalpy change by 4:
(PCl5(s) ΔH°rxn) / 4 = -1774.0 kJ / 4

Step 3: Calculate the enthalpy change for the formation of 1 mole of PCl5

Now we have the enthalpy change for the formation of 4 moles of PCl5. To calculate the enthalpy change for the formation of 1 mole of PCl5, we divide the previous result by 4:
(PCl5(s) ΔH°rxn) / 4 / 4 = -1774.0 kJ / 4 / 4

Step 4: Calculate the enthalpy change for the formation of 1.50 moles of PCl3

Since 1 mole of PCl5 is formed from 1 mole of PCl3, the enthalpy change for the formation of 1 mole of PCl3 will be the same as the enthalpy change for the formation of 1 mole of PCl5.

To calculate the enthalpy change for the formation of 1.50 moles of PCl3, we multiply the result from Step 3 by 1.50:
(PCl5(s) ΔH°rxn) / 4 / 4 x 1.50 = -1774.0 kJ / 4 / 4 x 1.50

Calculating this expression will give you the enthalpy change for the formation of 1.50 moles of PCl3 from phosphorus and chlorine.