Compute enthalpy for 2P(s) + 3Cl2(g) --> 2PCl3(l) at -639.4 kJ/mol to 2P(s) + 6Cl2(g) --> 2PCl5(s) at -887 kJ/mol to PCl5(l) + Cl2(g) --> PCl5(s) no given heat
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To compute the enthalpy change for a chemical reaction, you can use the concept of Hess's Law. According to Hess's Law, if a reaction can be expressed as a combination of other reactions, the overall enthalpy change of the reaction is equal to the sum of the enthalpy changes of those individual reactions.
In this case, we need to combine the given reactions to find the enthalpy change for the reaction 2P(s) + 3Cl2(g) → 2PCl3(l).
We have the following reactions:
1) 2P(s) + 6Cl2(g) → 2PCl5(s) ΔH = -887 kJ/mol
2) PCl5(l) + Cl2(g) → PCl5(s) ΔH = ? (unknown)
The reverse of reaction 2 is:
3) PCl5(s) → PCl5(l) + Cl2(g) ΔH = +x kJ/mol
Now, we can use the concept of reaction stoichiometry and Hess's Law to solve for the unknown ΔH value.
First, reverse reaction 1 to obtain the desired number of moles:
4) 2PCl5(s) → 2P(s) + 6Cl2(g) ΔH = +887 kJ/mol
Next, multiply reaction 2 by 2 to obtain the desired number of moles:
5) 2PCl5(l) + 2Cl2(g) → 2PCl5(s) ΔH = 2 * ΔH (reaction 2) = 2 * (-887) kJ/mol = -1774 kJ/mol
Using the given reactions, we can now manipulate the equations to cancel out the intermediate compounds:
Adding reactions 3 and 4 cancels out the PCl5(s) compound:
6) PCl5(l) + Cl2(g) + 2PCl5(s) → 2P(s) + 6Cl2(g) ΔH = x kJ/mol - 1774 kJ/mol
Since reaction 6 is the sum of reactions 3 and 4 (in the reverse direction), its enthalpy change will be equal to the sum of ΔH values for reactions 3 and 4:
ΔH for reaction 6 = x kJ/mol - 1774 kJ/mol
We also know that reaction 6 is the reverse of the desired reaction, so the enthalpy change for the desired reaction will be the negative of the value in reaction 6:
Enthalpy change for 2P(s) + 3Cl2(g) → 2PCl3(l) = -ΔH for reaction 6
Therefore, the enthalpy change for the reaction 2P(s) + 3Cl2(g) → 2PCl3(l) at -639.4 kJ/mol is:
-639.4 kJ/mol = -ΔH for reaction 6
ΔH for reaction 6 = 639.4 kJ/mol
To find the enthalpy change for the reaction PCl5(l) + Cl2(g) → PCl5(s) without a given heat, you need additional information or experimental data. Without a ΔH value for this reaction, it is not possible to determine the enthalpy change using Hess's Law alone.
I don't understand the question.
PCl5(l) + Cl2(g) --> PCl5(s) can't be correct.