3(20 pts) Consider the data: P4(s) + 6 Cl2(g) ----> 4 PC13(g) P4(s) + 5 O2(g)...
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
4. (10 points)Given the following data: P4 (s) + 6 Cl2 (g) -------> 4 PC13 (1) P4 (8) + 10 Cl2 (g) -------> 4 PCls (g) AG° = -1088 kJ/mole P4 AG° = -1292 kJ/mole P4 Calculate AGº for the reaction PC13 (1) + Cl2 (g) -------> PCI; (g) (answer only 1 point)
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
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 AH for the reaction PC13(1) + Cl2(g) → PC15(s) given the following data: Equation ΔΗ P4(s) + 6 Cl2(g) → 4 PC13(1) -1280 kJ P4(s) + 10 C12(g) → 4 PC1:($) -1774 kJ pt pt AH= 1 pt
age 1: Question 2 (34 points) Calculate A Hºrxn for the following reaction from the given data. PC15(g) + 4H2O(g) ---> H3PO4(1) + 5 HCl(g) A Hºrxn = ??? P4(s) + 10 Cl2(g) ---> 4 PC15(8) A Hºrxn = - 1499.6 kJ H2(g) + 1/2O2(g) ---> H2O(g) A Hºrxn = -241.8 kJ H2(g) + Cl2(g) ---> 2 HCl(g) A Hºrxn = -185.0 kJ 4 H3PO4(1) ---> P4010(s) + 6 H2O(g) A Hºrxn = +677.2 kJ : P4(s) + 5 O2()...
Given the heats of the following reactions: AH° kJ 1. P.(s) + 6C,(g) -->4PCI (g) -1225.6 II. P (s) + 50, (g) ---->P,(s) -2967.3 III. PCI, (g) + Cl, (g) ---> PCI (g) -84.2 IV. PCI,(g) + x0,(g)--->CI PO(g) -285.7 Calculate the value of AH® for the reaction below: P.O.(s) + 6PCI_(8) ---> 10C1, PO(g) -110.5 kJ -610.1 kJ -2682.2 kJ (D) -7555.0 kJ (A) (B)
Need Help 2. Given: Pa(s) + 3 O2(g) → P.O(s) AH = -1640.1 kJ Pa(s) + 5 O2(g) → P4010(s) AH = -2940.1 kJ Use Hess Law to calculate the AH for P.O(s) + 2O2(g)→ P4010(s)
Given: P4(s) + 3 O2(g) --> P406(s) ?H = -1640.1 kJ P4(s)+ 5 O2(g)--> P4O10(s). ?H=-2940.1 kJ Use Hess Law to calculate the ?H for P4O6(s) + 2O2(g)-->P4O10(s)
1) PC13(1) + Cl2(g) → PCI;(g) Use equations (2) and (3) to calculate AH,xn of equation (1): AH=-1280k (2) P4(5) + 6C12(3) — 4PCI3(0) (3) P4(s) + 10C12(g) → 4PC1g(8) AH = -1774k kJ