Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:
P4(g) + 10 Cl2(g) → 4PCl5(s), ΔH°rxn = ?
Given:
PCl5(s) → PCl3(g) + Cl2(g), ΔH°rxn= +157 kJ
P4(g) + 6 Cl2(g) → 4 PCl3(g), ΔH°rxn = -1207 kJ
-1835 kJ
To determine ΔH°rxn for the given reaction, we can use Hess's law, which states that the enthalpy change of a reaction can be calculated by summing the enthalpy changes of a series of reactions that add up to the desired reaction.
The given reactions are:
1) PCl5(s) → PCl3(g) + Cl2(g), ΔH°rxn = +157 kJ
2) P4(g) + 6 Cl2(g) → 4 PCl3(g), ΔH°rxn = -1207 kJ
To obtain the desired reaction:
P4(g) + 10 Cl2(g) → 4 PCl5(s)
where the coefficient of reaction 1 needs to be adjusted to produce 10 moles of Cl2, we multiply it by 5:
5 × (PCl5(s) → PCl3(g) + Cl2(g)) → 5 × (+157 kJ)
Now, we can write the adjusted reaction:
5PCl5(s) → 5PCl3(g) + 5Cl2(g)
Next, we need to multiply reaction 2 by 5 to produce 5 moles of PCl5:
5 × (P4(g) + 6 Cl2(g) → 4 PCl3(g)) → 5 × (-1207 kJ)
Finally, we can write the adjusted reaction:
5P4(g) + 30 Cl2(g) → 20 PCl3(g)
Now we can sum up the adjusted reactions:
5P4(g) + 30 Cl2(g) → 20 PCl3(g)
+ 5PCl5(s) → 5PCl3(g) + 5Cl2(g)
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5P4(g) + 35 Cl2(g) → 20 PCl3(g) + 5Cl2(g)
By canceling out the Cl2 terms on each side of the equation, we get:
5P4(g) + 35 Cl2(g) → 20 PCl3(g) + 5Cl2(g)
5P4(g) + 30 Cl2(g) → 20 PCl3(g)
The net reaction becomes:
P4(g) + 10 Cl2(g) → 4 PCl5(s)
Now we can sum up the corresponding ΔH°rxn values:
5 × (-1207 kJ) + 5 × (+157 kJ) = -6035 kJ + 785 kJ = -5249 kJ
Therefore, the value of ΔH°rxn for the given reaction is -5249 kJ.
To find ΔH°rxn for the reaction P4(g) + 10 Cl2(g) → 4PCl5(s), we can use Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps of the reaction.
Given the following reactions:
1) PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = +157 kJ
2) P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ
We can manipulate these reactions in order to match the desired reaction. Notice that reaction 1 has PCl5 as a product while reaction 2 has PCl3 as a product. We can multiply reaction 1 by 4 to obtain four moles of PCl5 so that we can cancel it out with reaction 2. We also need to multiply reaction 1 by 4 to obtain four moles of Cl2 to match the desired reaction.
4 * reaction 1: 4 PCl5(s) → 4 PCl3(g) + 4 Cl2(g) ΔH°rxn = 4 * (+157) kJ = +628 kJ
Now, we can sum up reaction 2 and the manipulated reaction 1:
P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = -1207 kJ
4 PCl5(s) → 4 PCl3(g) + 4 Cl2(g) ΔH°rxn = +628 kJ
By adding these reactions, we can cancel out PCl3 and Cl2 on both sides:
P4(g) + 10 Cl2(g) → 4 PCl5(s) ΔH°rxn = -1207 kJ + 628 kJ = -579 kJ
Therefore, the ΔH°rxn for the reaction P4(g) + 10 Cl2(g) → 4PCl5(s) is -579 kJ.