How do you find the change in heat/enthalpy?
P4 (s) + 5O2 (g) -----------> P4O10 (s)
The enthalpy change can be calculated from the standard enthalpy of formation of individual components. For this reaction, it is given by
dH= dHf(products) - dHf(reactants)
= dHf(P4O10) - dHf(P4) - 5dHf(O2)
= -2984-0-0 kJ/mol = -2984 kJ/mol
How do you find the change in heat/enthalpy? P4 (s) + 5O2 (g) -----------> P4O10 (s)
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
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
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
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)
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
White phosphorus (P4) is used to produce phosphorus oxides, such as P4O10. How many kilograms of P4O10 would be produced from 470. g of pure P4?
find enthalpy change
3. When 2.617 g of AX (s) dissolves in 141.2 g of water in a coffee-cup calorimeter the temperature rises from 24.5 °C to 36.5 °C. Calculate the enthalpy change (in kJ/mol) for the solution process. AX(s) - A+ (aq) + X(aq) Assumptions for this calculation: The specific heat of the solution is the same as that of pure water (4.18 J/gK) The density of water = 1.000 g/mL. The liquid's final volume is not changed by...
Given 2Al2O3 (s) --> 4Al(s) + 3O2 (g) (standard enthalpy change= 3351.4 kJ) a) What is the heat of formation of aluminum oxide? How do I find heat of formation from standard enthalpy change? I know how to do it when I'm given enthalpy, but they aren't the same thing- correct? Also, if this question gave me delta-H, I know that the heat of formation for 4Al = 0 and 3O2 = 0 because they are elementary species. This would...
Use Hess’s Law to find the standard enthalpy change for the reaction CO2(g) → C(s) + O2(g) using only the following information. Show all your work, including any equations you use to obtain your answer and showing clearly how you obtained that answer. (3 pts.) H2O(l) → H2(g) + 1/2O2(g) C2H6(g) → 2C(s)+ 3H2(g) 2CO2(g) + 3H2O(l) → C 2H6(g) + 7/2O2(g) ∆Ho (kJ) 643 kJ 190.6kJ 3511.1 kJ
The combustion of propane, C3H8, occurs via the reaction C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) with heat of formation values given by the following table: Substance ΔH∘f (kJ/mol) C3H8 (g)= -104.7 CO2(g)= −393.5 H2O(g)= −241.8 Calculate the enthalpy for the combustion of 1 mole of propane.