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(a)
Interpretation:
The ammonium carbonate partial pressure (Kp) should be calculated given equilibrium process and decomposition reaction.
Concept Introduction:
Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.
Kp: The equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is unites number, although it relates the pressures.
Thermal decomposition reaction: This reaction caused by heat or decomposition of starting substance is the temperature at which the substance chemically decomposes. In other words large molecule being broken down into single elements (or) compounds for example COCl2 readily decomposed into CO and Cl2.
To find: The equilibrium reaction should be identified given the statement.
(b)
Interpretation:
The ammonium carbonate partial pressure (Kp) should be calculated given equilibrium process and decomposition reaction.
Concept Introduction:
Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.
Kp: The equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is unites number, although it relates the pressures.
Thermal decomposition reaction: This reaction caused by heat or decomposition of starting substance is the temperature at which the substance chemically decomposes. In other words large molecule being broken down into single elements (or) compounds for example COCl2 readily decomposed into CO and Cl2.
To find: Calculate the partial pressure values (Kp) are given the decomposition reaction.
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Chapter 15 Solutions
Chemistry: Atoms First
- Nonearrow_forwardGiven the standard enthalpies of formation for the following substances, determine the reaction enthalpy for the following reaction. 4A (g) + 2B (g) → 2C (g) + 7D (g) AHrxn =?kJ Substance AH in kJ/mol A (g) - 20.42 B (g) + 32.18 C (g) - 72.51 D (g) - 17.87arrow_forwardDetermine ASran for Zn(s) + 2HCl(aq) = ZnCl2(aq) + H2(aq) given the following information: Standard Entropy Values of Various Substance Substance So (J/mol • K) 60.9 Zn(s) HCl(aq) 56.5 130.58 H2(g) Zn2+(aq) -106.5 55.10 CI (aq)arrow_forward
- 3) Catalytic hydrogenation of the compound below produced the expected product. However, a byproduct with molecular formula C10H12O is also formed in small quantities. What is the by product?arrow_forwardWhat is the ΔHorxn of the reaction? NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) ΔHorxn 1= ________ kJ/molarrow_forward= +92kJ ΔΗ = +170kJ Use the following reactions: 2NH3(9) N2(g) + 3H2(g) → 11/N2(g) + 2H2O (1) → NO2(g) + 2H2(g) Determine the DH° of this reaction: NO2(g) + H2(g) → 2(g) → 2H2O(l) + NH3(9) ΔΗarrow_forward
- Determine the entropy change for the reaction SO2(g) + O2(g) following information: Standard Entropy Values of Various Substance Substance SO2(g) 02(g) SO3(g) So (J/mol K) 248.2 205.0 256.8 → SO3(g) given thearrow_forwardIndicate which one of the following reactions most certainly results in a negative AS sys. O1402(g) + 3NH4NO3 (s) + C10 H22(1) → 3N2(g) + 17H2O(g) + 10CO2(g) ○ CO2(aq) = CO2(g) ○ H₂O(g) = H₂O(s) CaCO3(g) = CaO(s) + CO2(g) O CuSO4.5H2O(s) = CuSO4(s) + 5H2O(g)arrow_forwardEstimate the DH°rxn of the reaction below: H H-C-C=C-H H Н A table of bond energy Bond H Bond Energy (kJ/mol) C-H 413 C-O 360 C=O 743 C-C 348 |C = C 612 O-H 463 H-H 436 + H-H -> H H-C. - H | | 1 HHHarrow_forward
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