You are given the following data.
P4(s) + 6 Cl2(g) 4 PCl3(g) H = -1225.6 kJ
P4(s) + 5 O2(g) P4O10(s) H = -2967.3 kJ
PCl3(g) + Cl2(g) PCl5(g) H = -84.2 kJ
PCl3(g) + 1/2 O2(g) Cl3PO(g) H = -285.7 kJ
Calculate H for the following reaction.
P4O10(s) + 6 PCl5(g) 10 Cl3PO(g)
Do you omit the arrows on purpose? I can't tell reactants from products without the arrows. Please repost and show the arrows.
You are given the following data.
P4(s) + 6 Cl2(g)--> 4 PCl3(g) H = -1225.6 kJ
P4(s) + 5 O2(g) --> P4O10(s) H = -2967.3 kJ
PCl3(g) + Cl2(g) --> PCl5(g) H = -84.2 kJ
PCl3(g) + 1/2 O2(g) --> Cl3PO(g) H = -285.7 kJ
Calculate H for the following reaction.
P4O10(s) + 6 PCl5(g) --> 10 Cl3PO(g)
i got -610.1kJ is that correct?
I obtained the same number of 610.1 kJ.
and it was a positive, right?
No. I think it is a negative 610.1 kJ.
Equations 2 and 3 are reversed. Equations 1 and 4 are as written. Equation 3 is multiplied by 6 and equation 4 is multiplied by 10. Check my numbers.
yep you are right thanks
To calculate the value of ΔH for the given reaction, we can use the concept of Hess's Law. Hess's Law states that the total enthalpy change of a reaction is independent of the pathway taken to reach the products and is equal to the sum of the individual enthalpy changes of the intermediate reactions.
1. Start by examining the given reaction equation:
P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g)
2. Look for any intermediate reactions that involve the same compounds as the reactants or products in the given reaction. In this case, we have the following intermediate reactions:
P4(s) + 5 O2(g) → P4O10(s) (Reaction 1)
PCl3(g) + 1/2 O2(g) → Cl3PO(g) (Reaction 2)
PCl3(g) + Cl2(g) → PCl5(g) (Reaction 3)
3. We need to adjust the coefficients of the intermediate reactions to match the coefficients in the given reaction equation. To achieve this, we can manipulate the intermediate reactions as follows:
2 x (Reaction 1): 2 P4(s) + 10 O2(g) → 2 P4O10(s)
20 x (Reaction 2): 20 PCl3(g) + 10 O2(g) → 20 Cl3PO(g)
12 x (Reaction 3): 12 PCl3(g) + 6 Cl2(g) → 12 PCl5(g)
4. Now, we can sum up the adjusted intermediate equations to obtain the overall reaction equation:
2 P4(s) + 10 O2(g) + 20 PCl3(g) + 6 Cl2(g) → 2 P4O10(s) + 20 Cl3PO(g) + 12 PCl5(g)
5. Finally, we can calculate the overall enthalpy change (ΔH) for the given reaction by summing the enthalpy changes of the adjusted intermediate reactions:
ΔH = 2 x (Reaction 1) + 20 x (Reaction 2) + 12 x (Reaction 3)
Note: The enthalpy changes for the intermediate reactions are given in the question.
ΔH = 2 x (-2967.3 kJ) + 20 x (-285.7 kJ) + 12 x (-84.2 kJ)
ΔH = -5946.6 kJ + (-5714 kJ) + (-1010.4 kJ)
ΔH ≈ -12671 kJ
Therefore, the enthalpy change (ΔH) for the given reaction P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g) is approximately -12671 kJ.