For each problem, find the ΔH for the target reaction below, given the following step reactions and subsequent ΔH values:
Target Reaction: PCl5(g) → PCl3(g) + Cl2(g)
Step Reactions: P4 (s) + 6 Cl2 (g) → 4 PCl3 (g) ΔH = -2439 kJ
4 PCl5 (g) → P4 (s) + 10 Cl2 (g). ΔH = 3438 kJ
Add the two equations to obtain the following:
P4 (s) + 6 Cl2 (g) → 4 PCl3 (g)
4 PCl5 (g) → P4 (s) + 10 Cl2 (g).
---------------------------------------------------
P4 (s) + 6 Cl2 (g) + 4 PCl5 (g) ==> 4 PCl3 (g) + P4 (s) + 10 Cl2 (g)
Now cancel P4 on left and right. 6 Cl2 on left and 6 Cl2 on right leaving 4 Cl2 on right. This leaves:
4PCl5 ==> 4PCl3 + 4 Cl2 with dH being the sum of the two (-2439 + 3438 = ?)
That reaction is just four times the target reaction so divide the final reaction by 4 to obtain PCl5 ==> PCl3 + Cl2 and of course divide the sum above by 4 to get delta H for the target reaction.
To find the ΔH for the target reaction, we need to use Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps.
Given the step reactions and their corresponding ΔH values, we can use them to calculate the ΔH for the target reaction.
Step 1: P4 (s) + 6 Cl2 (g) → 4 PCl3 (g) ΔH = -2439 kJ
Step 2: 4 PCl5 (g) → P4 (s) + 10 Cl2 (g). ΔH = 3438 kJ
To get the target reaction, we need to reverse the first step reaction and multiply it by 4 so that the PCl3 cancels out:
Step 1 (reversed and multiplied by 4): 4 PCl3 (g) → P4 (s) + 24 Cl2 (g) ΔH = 4 × (-2439 kJ) = -9756 kJ
Now we have the step reactions and their corresponding ΔH values. To find the ΔH for the target reaction, we add the ΔH values of the step reactions:
ΔH for the target reaction = ΔH (Step 2) + ΔH (Step 1 reversed and multiplied by 4)
ΔH for the target reaction = 3438 kJ + (-9756 kJ)
= -6318 kJ
Therefore, the ΔH for the target reaction PCl5(g) → PCl3(g) + Cl2(g) is -6318 kJ.
To find the ΔH for the target reaction, we need to consider the step reactions and their ΔH values.
Step Reaction 1: P4 (s) + 6 Cl2 (g) → 4 PCl3 (g) ΔH = -2439 kJ
Step Reaction 2: 4 PCl5 (g) → P4 (s) + 10 Cl2 (g). ΔH = 3438 kJ
Since the target reaction involves the formation of PCl3 and Cl2 from PCl5, we need to reverse Step Reaction 1 and multiply its ΔH by -1:
-1 * (-2439 kJ) = 2439 kJ
Next, we need to adjust the coefficients of Step Reaction 2 to match the target reaction. Since we need 1 mole of PCl3 and 1 mole of Cl2 in the target reaction, we can use a factor of 1/4:
(1/4) * (3438 kJ) = 859.5 kJ
Adding these two values together gives us the ΔH for the target reaction:
ΔH = 2439 kJ + 859.5 kJ = 3298.5 kJ
Therefore, the ΔH for the target reaction PCl5(g) → PCl3(g) + Cl2(g) is 3298.5 kJ.