Assume the reaction of 60.63 g of Cl2 in the following reaction to determine the other quantities.
| 6Cl2 + 1P4 -> 4PCl3 | |
|---|---|
| moles of Cl2 reacting = | ____mol |
| moles of P4 required = | _____mol |
| moles of PCl3 formed = | _____mol |
| mass of PCl3 formed = | ______ g |
Assume the reaction of 60.63 g of Cl2 in the following reaction to determine the other...
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
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
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
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
Determine the value of the equilibrium constant, Kgoal, for the reaction N2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=? by making use of the following information: 1. N2(g)+O2(g)⇌2NO(g), K1 = 4.10×10−31 2. N2(g)+2H2(g)⇌N2H4(g), K2 = 7.40×10−26 3. 2H2O(g)⇌2H2(g)+O2(g), K3 = 1.06×10−10 Express your answer numerically. Kgoal = 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. Kgoal =
What is the value of delta G rxn for the reaction 4PCl3 (g) -> 6 Cl2(g) + P4(g) Delta G / kJ * mol - PCl3(g): -296.6, P4(g): 24.1 A) -1103 kJ B) -1054 kJ C) -321 kJ D) 1103 kJ
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.
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.
Elemental phosphorus reacts with chlorine gas according to the equation P4(s)+6Cl2(g)→4PCl3(l) A reaction mixture initially contains 45.39 g P4 and 130.4 g Cl2. Once the reaction has reached completion, what mass (in g) of the excess reactant is left?
Part ADetermine the value of the equilibrium constant, Kgoal , for the reactionN2(g)+O2(g)+H2(g)⇌12N2H4(g)+NO2(g) , Kgoal=?by making use of the following information:1. N2(g)+O2(g)⇌2NO(g) , K1 =4.10×10−312. N2(g)+2H2(g)⇌N2H4(g) , K2 =7.40×10−263. 2NO(g)+O2(g)⇌2NO2(g) , K3 =6.00×10−13Express your answer numerically.Part BDetermine the equilibrium constant, Kgoal , for the reaction4PCl5(g)⇌P4(s)+10Cl2(g), Kgoal=?by making use of the following information:P4(s)+6Cl2(g)⇌4PCl3(g), K1=2.00×1019PCl5(g)⇌PCl3(g)+Cl2(g), K2=1.13×10−2Express your answer numerically.