How would you calculate delta H for the following reaction:
2P + 5Cl2 -> 2PCl5
Given:
PCl3 + Cl2 -> PCl5 Delta H=-87.9 kJ
2P + 3Cl2 -> 2PCl3 Delta H=-574 kJ
I have no clue how to do this since I was absent. Can someone help?
To calculate the delta H for the given reaction, you can use Hess's Law. Hess's Law states that the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps of the reaction, provided the reactants and products are the same.
Here's the step-by-step process to calculate delta H for the reaction:
1. Write down the given reactions and their respective delta H values:
a. PCl3 + Cl2 -> PCl5 ΔH = -87.9 kJ (given)
b. 2P + 3Cl2 -> 2PCl3 ΔH = -574 kJ (given)
c. 2P + 5Cl2 -> 2PCl5 ΔH = ?
2. Note that reaction (a) needs to be reversed because the given reaction is the reverse of reaction (a). Reverse reaction (a) and its delta H:
PCl5 -> PCl3 + Cl2 ΔH = +87.9 kJ
3. Multiply reaction (b) by 2 to balance the number of P and Cl2 atoms with reaction (c):
4P + 6Cl2 -> 4PCl3 ΔH = -1148 kJ
4. By comparing reactions (a), (b), and (c), you can see that reaction (c) is equal to reaction (a) + reaction (b):
4P + 6Cl2 + PCl5 -> 4PCl3 + PCl5 -> 4PCl3 + Cl2 + PCl5 -> 2PCl5 ΔH = (-1148 kJ) + (+87.9 kJ) = -1060.1 kJ
Therefore, the delta H for the reaction 2P + 5Cl2 -> 2PCl5 is -1060.1 kJ.
To calculate the change in enthalpy (ΔH) for the reaction, "2P + 5Cl2 -> 2PCl5", you can use Hess's Law, which states that the total enthalpy change of a reaction is independent of the route taken. By considering the given reactions, you can manipulate them to obtain the desired reaction and, at the same time, manipulate their enthalpy changes. Here's how to do it step-by-step:
1. Reverse the equation "PCl3 + Cl2 -> PCl5" and change the sign of the enthalpy change:
PCl5 -> PCl3 + Cl2 ΔH = +87.9 kJ
2. Multiply the reversed equation by 2 to obtain the same number of moles of PCl5 as in the desired reaction:
2PCl5 -> 2PCl3 + 2Cl2 ΔH = +2 * 87.9 kJ = +175.8 kJ
3. Multiply the equation "2P + 3Cl2 -> 2PCl3" by 2 to balance the number of moles of PCl3:
4P + 6Cl2 -> 4PCl3 ΔH = +2 * (-574 kJ) = -1148 kJ
4. Add the two modified equations together to obtain the desired reaction:
2PCl5 + 4P + 6Cl2 -> 2PCl3 + 2Cl2 + 4PCl3 + 2Cl2
Simplify the equation:
2PCl5 + 4P + 6Cl2 -> 6PCl3 + 4Cl2
ΔH = +175.8 kJ + (-1148 kJ) = -972.2 kJ
Therefore, the change in enthalpy (ΔH) for the reaction "2P + 5Cl2 -> 2PCl5" is -972.2 kJ.
Look up Hess' Law.
Multiply equation 1 by 2 and add it to equation 2. Multiply delta H for equation 1 by 2 and add to delta H for equation 2.