You are given the following data.
P4(s) + 6 Cl2(g) 4 PCl3(g) ΔH = -1225.6 kJ
P4(s) + 5 O2(g) P4O10(s) ΔH = -2967.3 kJ
PCl3(g) + Cl2(g) PCl5(g) ΔH = -84.2 kJ
PCl3(g) + 1/2 O2(g) Cl3PO(g) ΔH = -285.7 kJ
Calculate ΔH for the following reaction.
P4O10(s) + 6 PCl5(g) 10 Cl3PO(g)
Well, first things first, let's get our reaction straight. We have P4O10(s) and 6 PCl5(g) reacting to form 10 Cl3PO(g).
Now, let's see how we can use the given data to calculate ΔH for this reaction.
We can break down the reaction into a series of steps using Hess's law.
Step 1: P4(s) + 5 O2(g) -> P4O10(s) with a ΔH of -2967.3 kJ.
Step 2: P4(s) + 6 Cl2(g) -> 4 PCl3(g) with a ΔH of -1225.6 kJ.
Step 3: PCl3(g) + Cl2(g) -> PCl5(g) with a ΔH of -84.2 kJ.
Step 4: PCl3(g) + 1/2 O2(g) -> Cl3PO(g) with a ΔH of -285.7 kJ.
To calculate ΔH for the overall reaction, we need to manipulate these equations and add them up in a way that cancels out any substances that appear and disappear in different steps.
So, let's rearrange Step 2 by multiplying it by 2 to obtain 2 PCl3(g):
2(P4(s) + 6 Cl2(g) -> 4 PCl3(g)) with a ΔH of -2(1225.6 kJ) = -2451.2 kJ.
Now, we can add Step 1, Step 3, and the modified Step 2 to obtain the overall reaction:
P4O10(s) + 2 PCl3(g) + 2 PCl5(g) + 1/2 O2(g) -> 2 PCl5(g) + 10 Cl3PO(g).
The ΔH for the overall reaction can be calculated by adding up the respective ΔH values for each step:
-2967.3 kJ + (-2451.2 kJ) + (-84.2 kJ) + (-285.7 kJ) = -6788.4 kJ.
So, the ΔH for the overall reaction is -6788.4 kJ.
And that's how you get the answer. Just be sure to double-check your calculations - you don't want to end up with a clownish mistake!
To calculate ΔH for the given reaction, you need to use the Hess's law, which states that the total enthalpy change for a reaction is equal to the sum of the enthalpy changes of the individual steps involved in the reaction.
Let's break down the given reaction into individual steps using the given data:
Step 1: P4(s) + 5 O2(g) → P4O10(s) ΔH = -2967.3 kJ (Given)
Step 2: PCl3(g) + 1/2 O2(g) → Cl3PO(g) ΔH = -285.7 kJ (Given)
Step 3: P4(s) + 6 Cl2(g) → 4 PCl3(g) ΔH = -1225.6 kJ (Given)
Step 4: PCl3(g) + Cl2(g) → PCl5(g) ΔH = -84.2 kJ (Given)
Now, let's rearrange the equations to obtain the desired reaction:
Desired reaction: P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g)
To obtain this reaction, we'll need to manipulate the given equations and multiply them by appropriate factors:
Step 1 (multiplied by 2): 2P4(s) + 10 O2(g) → 2P4O10(s) ΔH = -2 * (-2967.3 kJ) = +5934.6 kJ
Step 2: PCl3(g) + 1/2 O2(g) → Cl3PO(g) ΔH = -285.7 kJ (Given)
Step 3 (multiplied by 3): 3P4(s) + 18 Cl2(g) → 12 PCl3(g) ΔH = -3 * (-1225.6 kJ) = +3676.8 kJ
Step 4 (multiplied by 6): 6PCl3(g) + 6Cl2(g) → 6PCl5(g) ΔH = -6 * (-84.2 kJ) = +505.2 kJ
Now, let's combine the individual steps to calculate the ΔH for the desired reaction:
P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g)
ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4
= 5934.6 kJ + (-285.7 kJ) + 3676.8 kJ + 505.2 kJ
= 9829.9 kJ
Therefore, the ΔH for the given reaction is 9829.9 kJ.
To calculate ΔH for the given reaction, P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g), you can apply the concept of Hess's Law. Hess's Law states that the enthalpy change of a reaction is the same, regardless of the route taken from reactants to products, as long as the initial and final conditions are the same.
Here's how to calculate ΔH:
1. First, determine the steps needed to reach the desired reaction from the given data. We can see that the given reactions have some of the same reactants and products as the desired reaction. We need to manipulate the given reactions in a way that aligns with the desired reaction.
2. Observe that the reaction P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g) can be obtained by combining the following two reactions, along with necessary multiplication factors to balance them:
i. P4(s) + 5 O2(g) → P4O10(s) (Given ΔH = -2967.3 kJ)
ii. PCl3(g) + 3/2 Cl2(g) → 3/2 PCl5(g) (Given ΔH = -84.2 kJ)
3. Multiply reaction i by 6 and reaction ii by 4 to obtain the desired reaction equation:
6P4(s) + 30O2(g) → 6P4O10(s) (Multiply reaction i by 6)
4PCl3(g) + 6Cl2(g) → 6PCl5(g) (Multiply reaction ii by 4)
Combining both modified reactions, you get:
6P4(s) + 30O2(g) + 4PCl3(g) + 6Cl2(g) → 6P4O10(s) + 6PCl5(g)
4. Now, you can add the given ΔH values of the original reactions to calculate the ΔH of the desired reaction:
ΔH = (6 × ΔH of reaction i) + (4 × ΔH of reaction ii)
= (6 × -2967.3 kJ) + (4 × -84.2 kJ)
= -17803.8 kJ - 336.8 kJ
= -18140.6 kJ
Therefore, ΔH for the reaction P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g) is -18140.6 kJ.