Calculate the value of ΔH° for the following reaction: P4O10(s) + 6PCl5(g) ---> 10Cl3PO(g)
P4(s) + 6Cl2(g) ---> 4PCl3(g) |
ΔH° = -1225.6 kJ |
P4(s) + 5O2(g) ---> P4O10(s) |
ΔH° = -2967.3 kJ |
PCl3(g) + Cl2(g) ---> PCl5(g) |
ΔH° = -84.2 kJ |
PCl3(g) + (1/2)O2(g) ---> Cl3PO(g) |
ΔH° = -285.7 k |
please explain well i do not understand these kind of problems
Calculate the value of ΔH° for the following reaction: P4O10(s) + 6PCl5(g) ---> 10Cl3PO(g) P4(s) +...
Assume all reactants and products are gases unless noted otherwise. Given the following data; P4(s) + 6Cl2 → 4PCl3 ∆H = -1225.6 kJ P4(s) + 5O2 → P4O10(s) ∆H = -2967.3 kJ PCl3 + Cl2 → PCl5 ∆H = -84.2 kJ PCl3 + 1/2O2 → Cl3PO ∆H = -285.7 kJ Calculate ∆Hrxn for the following reaction. P4O10 + 6PCl5 → 10Cl3 PO
32. Given the following data: A. P4(s)+ 6 Cl2(g)4 PCls(g) B. P4(s)+ 5 02(g) P4O10(s)AH=- 2967.3 C. PCls(g)+Cl2(g) PCIS(g) D. PCI3(g)+O2(g) ClaPO(g) AH-285.7 AH 1225.6 kJ AH=-84.2 Calculate AH for the reaction: P4O10(s)+6 PCI5(g) 10 CI3PO(g). Note when I solved this problem, I used 3 of the 4 data reactions, added them together, then used the fourth data reaction to get to the desired reaction. -B-6C+10D+A
Calculate the standard entropy change for the reaction P4(g)+5O2(g)→P4O10(s) using the data from the following table: Substance ΔH∘f (kJ/mol) ΔG∘f (kJ/mol) S∘ [J/(K⋅mol)] P4(g) 58.9 24.5 279.9 O2(g) 0 0 205.0 P4O10(s) −2984 −2698 228.9
Determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)⇌P4(s)+10Cl2(g), Kgoal=? by making use of the following information: P4(s)+6Cl2(g)⇌4PCl3(g), K1=2.00×1019 PCl5(g)⇌PCl3(g)+Cl2(g), K2=1.13×10−2
3(20 pts) Consider the data: P4(s) + 6 Cl2(g) ----> 4 PC13(g) P4(s) + 5 O2(g) -----> P4010(s) PC13(g) + Cl2(g) ---> PCls(g) PC13(g) + 1/2O2(g) ----> C13PO(g) AH° (kJ/mol) -1226.6 -2967.3 -84.2 -285.7 Using only the above reaction data, compute the AHX of P4010(s) + 6 PCls(8) ----> 10 C13PO (g)
Express the equilibrium constant for the following reaction. P4(s) + 5O2(g) ⇌ P4O10(aq) K =___ A. A) = [P4O10] / [P4] B. B) = [P4O10] / [P4][O2]5 C. C) = [P4O10] / [O2]5 D. D) [O2]-5 E. E) none of these
Given: PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = + 157 kJ P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = - 1207 kJ What is the standard-state enthalpy change for the following reaction? P4(g) + 10 Cl2(g) → 4PCl5(s)a -2100 kJb-1835 kJc-1364 kJD -1786 kJ
Assume the reaction of 118.30 g of P4 in the following reaction to determine the other quantities. 1P4 + 6Cl2 -> 4PCl3 moles of P4 reacting = mol moles of Cl2 required = mol moles of PCl3 formed = mol mass of PCl3 formed = g
For the reaction 4PCl3(g)⇌P4(s)+6Cl2(g) KA= 240.7 The reverse reaction is P4(s)+6Cl2(g)⇌4PCl3(g) What is the value for Kreverse? Enter a numerical answer only.
Part A Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the following information: 1. 2CO2(g)+2H2O(l)⇌CH3COOH(l)+2O2(g), K1 = 5.40×10−16 2. 2H2(g)+O2(g)⇌2H2O(l), K2 = 1.06×1010 3. CH3COOH(l)⇌2C(s)+2H2(g)+O2(g), K3 = 2.68×10−9 Express your answer numerically. Part B Determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)⇌P4(s)+10Cl2(g), Kgoal=? by making use of the following information: P4(s)+6Cl2(g)⇌4PCl3(g), K1=2.00×1019 PCl5(g)⇌PCl3(g)+Cl2(g), K2=1.13×10−2 Express your answer numerically.