Given these two reactions, how do I calculate
ΔH for the third one?
P4(s) + 6 Cl2(g) -> 4 PCl3(g) ΔH= -1148 kJ
P4(s) + 10 Cl2(g) -> 4 PCl5(g) ΔH = -1500 kJ
PCl3(g) + Cl2(g) -> PCl5(g) ΔH = ?
look up hess's law. you should be able to figure it out
Reverse equation 1 and add to equation and that gives you
4PCl3 + 4Cl2 ==> 4PCl5 which is just 4x what yu want so divide everything by 4.
To calculate ΔH for the third reaction, you can use the concept of Hess's law. Hess's law states that the enthalpy change of a reaction is independent of the pathway taken from the initial to the final state and is solely determined by the initial and final states of the reaction. In other words, you can obtain the enthalpy change of a reaction by combining other reactions with known enthalpy changes.
In this case, you need to find a combination of the given reactions that can cancel out the reactants and products in the desired reaction equation, leaving only the desired reaction and its enthalpy change.
Here's how you can do it:
1. Multiply the first equation by 2 to make the number of moles of PCl5 equal to that in the desired reaction equation:
2(P4(s) + 6 Cl2(g) -> 4 PCl3(g)) ΔH = -2 * 1148 kJ = -2296 kJ
2. Multiply the second equation by -4 to make the number of moles of PCl3 equal to that in the desired reaction equation:
-4(P4(s) + 10 Cl2(g) -> 4 PCl5(g)) ΔH = 4 * (-1500 kJ) = -6000 kJ
3. Flip the first equation and multiply it by 10 to make the number of moles of Cl2 equal to that in the desired reaction equation:
10 (4 PCl3(g) -> P4(s) + 6 Cl2(g)) ΔH = 10 * (-1148 kJ) = -11480 kJ
4. Add the three equations together to obtain the desired reaction and its enthalpy change:
-2296 kJ - 6000 kJ - 11480 kJ = -19776 kJ
Therefore, the enthalpy change for the reaction PCl3(g) + Cl2(g) -> PCl5(g) is -19776 kJ.