Concept explainers
(a)
Interpretation:
The mole fraction of product composition to be calculated. Also, the percent conversion of
Concept introduction:
For a single reaction system, the final moles of each of the components present, can be estimated by the equation:
Here,
Mole fraction
Here,
(b)
Interpretation:
By neglecting the effect of pressure on enthalpies, the amount of heat added or removed from the process should be determined.
Concept introduction:
The amount of heat transferred is,
Here,
The enthalpies of reactants and products depends on the heat capacity and temperature.
(c)
Interpretation:
The extent of reaction along with the amount of heat removed with the values of temperatures ranging from
Concept introduction:
The amount of heat transferred is,
Here,
The enthalpies of reactants and products depends on the heat capacity and temperature.
(d)
Interpretation:
The effect of pressure on the reaction with the help of extent of conversion and rate of heat transfer at 1 MPa and 15 MPa should be determined.
Concept introduction:
For a single reaction system, the final moles of each of the components present, can be estimated by the equation:
Here,
Mole fraction
Here,
(e)
Interpretation:
The purpose of selecting initial condition of
Concept introduction:
Adiabatic flame temperature helps to determine the completion of process. Adiabatic flame temperature occurs in two phases that is constant volume process and constant pressure process.
This temperature is applicable when the process takes place in an adiabatic condition which means no exchange of heat with the surrounding.
For a single reaction system, the final moles of each of the components present, can be estimated by the equation:
Here,
Mole fraction
Here,
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Chapter 9 Solutions
EBK ELEMENTARY PRINCIPLES OF CHEMICAL P
- For the following combustion reactions, predict whether compression will favour the formation of products or reactants, or whether there will be no change in the equilibrium composition:(a) C4H8(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(I)(b) C2H5OH(I) + 3(1)/(2) O2(g) → 2 CO2(g) + 3 H2O(I)(c) C6H12O11(s) + 3(1)/(2) O2(g) → 6 CO2(g) + 6 H2O(I)(d) 2 C6H5NH2(1) + 17(1)/(2) O2(g) → 12 CO2(g) + 7 H2O(I) +2 NO2(g)arrow_forwardThe reaction N2O4−⇀↽−2NO2 is allowed to reach equilibrium in a chloroform solution at 25 ∘C . The equilibrium concentrations are 0.327 mol/L N2O4 and 1.91 mol/L NO2 . Calculate the equilibrium constant, Kc , for this reaction. Kc= 11.16 An additional 1.00 mol NO2 is added to 1.00 L of the solution and the system is allowed to reach equilibrium again at the same temperature. Select the direction of the equilibrium shift after the NO2 is added. towards the product towards the reactant no change Determine how the addition of extra NO2 in the previous step will affect the rate constant, Kc . Kc will increase Kc will not change Kc will decrease Calculate the equilibrium concentrations of N2O4 and NO2 after the extra 1.00 mol NO2 is added to 1.00 L of solution. [N2O4]= mol/L [NO2]= mol/Larrow_forwardIn the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at 800°C to give hydrogen and carbon monoxide: CH4 (g) + H2O (g) ⇌ CO (g) + 3H2 (g) ΔH°=260 kJ/mol The secondary stage is carried out at about 1000°C, in the presence of air, to convert the remaining methane to hydrogen: CH4 (g) + 1/2 O2 (g) ⇌ CO (g) + 2 H2 (g) ΔH°=35.7 kJ/mol (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stage? (b) The equilibrium constant Kc for the primary stage is 18 at 800°C. Calculate Kp for the reaction. (c) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium? Thank you so much!arrow_forward
- For the reaction H₂(g) + Cl₂(g) 2 HCI(g), the equilibrium constant Kat 800°C is 4.35 x 104. Hydrogen and chlorine, each at a partial pressure of 0.700 bar, are placed in a vessel at 800°C and allowed to equilibrate. Find the final partial pressures of all three gases in this reaction. p(H₂) p(Cl₂)= i p(HCI) = i bar bar bararrow_forwardSolid NH4I ammonium iodide (which is always considered to be in excess) is introduced into an initially empty container at a constant temperature of 427°C. An equilibrium is quickly established and the total pressure (PT) is then 0.9 bar. NHẠI (s) 2 NH3(g) + HI(g) Express the equilibrium constant K' of this reaction as a function of PT a. Express the equilibrium constant K' of this reaction as a function of PT- The answer is PNH, PHI - - ( 2 2 P K' || po2 p°2 2P°, Can you explain why we get (p/2)^2?arrow_forwardThe system described by the reaction CO(g) + Cl2(g) = COCl2(g) is at equilibrium at a given temperature when Pco=0.30bar, Pcl2=0.10bar, and Pcocl2=0.60 bar. An additional pressure of Cl2(g)=0.40 bar is added. Find the pressure of CO when the system returns to equilibrium.arrow_forward
- Kc for the reaction CO2(g) + H2(g) ⇌ H2O(g) + CO(g) is 1.6 at about 990°C. Calculate the number of moles of water in the final equilibrium system obtained by initially adding 1.00 mol of H2, 2.00 mol of CO2, 0.750 mol of H2O, and 1.00 mol of CO to a 5.00 L reactor at 990°C.arrow_forward30. The following reaction takes place. A,B(s) = 2A(ac) + B(ag) + heat. If 1.80 moles of A,B is added into a 1.00 L container, it is found that 30.0% of the reactant dissociates to reach equilibrium. Find the solubility product constant, Ksp, for this reaction. Note: Your answer must include an ICE table. (aq),arrow_forwardHCH₂CO₂(aq)+CH, NH (aq)-CH₂CO₂(aq)+CH, NH(aq) At the temperature the engineer picks, the equilibrium constant K for this reaction is 0.86. The engineer charges ("fills") four reaction vessels with acetic acid and methylamine, and lets the reaction begin. He then measures the composition mixture inside each vessel from time to time. His first set of measurements are shown in the table below. Predict the changes in the compositions the engineer should expect next time he measures the compositions. reaction vessel B compound HCH,CO CHÍNH, CH,CO. CHÍNH HCH,CO, CHÍNH, CH,CÓ, CHÍNH, HCH.CO. concentration 0.89 M 0.61 M 0.69 M 0.68 M 0,43 M 0.15 M 1.15 M 1.14 M LILM expected change in concentration increase O decrease Ot increase O decrease O increase O decrease O increase O decrease O1 increase O decrease O1 increase O decrease Of increase O decrease Of increase O decrease (no change) O (no change) O (no change) (no change) O(no change) O (no change) O (no change) O (no change)arrow_forward
- N₂(9)+ 3H₂(g) → 2NH₂(9) At the temperature the engineer picks, the equilibrium constant K, for this reaction is 0.0069. The engineer charges ("fills") three reaction vessels with nitrogen and hydrogen, and lets the reaction begin. He then measures the composition of the mixture inside each vessel from time to time. His first set of measurements are shown in the table below. Predict the changes in the compositions the engineer should expect next time he measures the compositions. reaction vessel B compound N₂ 16₂ NH, N₂ H₂ E NH, N₂ H₂ NH, pressure 32.93 am 21.78 atm 43.94 atm 37.26 am 23.66 $8.50 m 32.56 atm 20.68 atm 44.67 an expected change in pressure O increase O increase O increase O increase O Increase Ot increase O decrease O decrease decrease O Increase O increase O O Increase O decrease O decrease Odecrease O decrease decrease decrease O(no change) O (no change) O(no change) O(no change) O (no change) (no change) (no change) O (no change) (no change)arrow_forwardThe production of 1,3-butadiene can be carried out by the dehydrogenation of n-butane: C 4H 10 (9) H 2C=CHHC=CH 2 (g) + 2 H 2 (g) The reaction is carried out at 925 K and 0.5 bar pressure. Starting with pure n-butane gas it comes to equilibrium. At 925 K the equilibrium constant (K) of the reaction is 0.30. Assuming ideal gases, calculate the fractional conversion of n-butane at equilibrium. 0.29 0.41 0.52 0.63 0.71 0.85 OOarrow_forwardA closed 1.00 L system initially containing 0.00200 mol H₂ and 0.00500 mol 1₂ at a given temperature is allowed to reach equilibrium Analysis of the equilibrium mixture shows that the amount of HI present at equilibrium is 0.00371 mol. Calculate the Kc at this temperature for the reaction. State whether the equilibrium is product-favored or reactant- favored at this temperature. H₂(g) +1₂(g) 2H1(g) Kc 31: The equilibrium is product-favored. Kc 31: The equilibrium is reactant-favored. Kc 0.032; The equilibrium is reactant-favored. Kc - 0.032; The equilibrium is product-favored.arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning