Equilibrium
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7.20. One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.
FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and pCO2= 0.80 atm?
7.20. One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.
FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and pCO2= 0.80 atm?
The initial partial pressures of CO and CO2 are 1.40 atm and 0.8. atm respectively.
The expression for the reaction quotient is:
Qp=PCO2 / PCO=0.80 / 1.4
Upvote 
7.19. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was reached, the concentration of PCl5 was found to be 0.5 x 10-1 mol L-1. If Kc is 8.3 x 10-3 what are the concentrations of PCl3 and Cl2 at equilibrium?
7.19. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was reached, the concentration of PCl5 was found to be 0.5 x 10-1 mol L-1. If Kc is 8.3 x 10-3 what are the concentrations of PCl3 and Cl2 at equilibrium?
Let the initial molar concentration of PCl5 per litre = x mol
Molar concentration of PCl5 at equilibrium = 0.05 mol
Therefore, moles of PCl5 decomposed = (x – 0.05) mol
Moles of PCl3 formed = (x – 0.05) mol
Moles of Cl2 formed = (x – 0.05) mol
The molar conc./litre of reactants a
Upvote 7.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
CH3COOH (l) + C2H5OH (l) ⇌CH3COOC2H5 (l) + H2O (l)
(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
7.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
CH3COOH (l) + C2H5OH (l) ⇌CH3COOC2H5 (l) + H2O (l)
(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
(i) The concentration ratio (Concentration quotient), Qc for the reaction is:
Qc = [CH3COOC2H5] [ H2O] / [CH3COOH] [C2H5OH]
(ii)
| CH3COOH | C2H5OH | CH3COOC2H5 | H2O |
Initial molar concentration | 1.0 mol | 0.18 mol | 0 | 0 |
Molar concentration at equilibrium | (1 – 0.171) = 0.829 mol | (0.18 – 0.171) = 0.009 mol | 0.171 mol | 0.171 mol |
Applying
Kc = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]
= (0.171 mol) x (0.171 mol) / (0.829 mol) (0.009 mol)
= 3.92
(iii)
| CH3COOH | C2H5OH | CH3COOC2H5 | H2O |
Initial molar concentration | 1.0 mol | 0.5 mol | 0.214 mol | 0.214 mol |
Molar concentration at equilibrium | (1 – 0.214) = 0.786 mol | (0.5 – 0.214) = 0.286 mol | 0.214 mol | 0.214 mol
|
Qc = [CH3COOC2H5] [H2O] / [CH3COOH] [
Upvote 7.17. Kp =0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration ok C2H6 when it is placed in a flask at 4 atm pressure,and allowed to come to equilibrium.
C2H6(g) ⇌ C2H4(g) + H2(g).
7.17. Kp =0.04 atm at 898 K for the equilibrium shown below. What is the equilibrium concentration ok C2H6 when it is placed in a flask at 4 atm pressure,and allowed to come to equilibrium.
C2H6(g) ⇌ C2H4(g) + H2(g).
The equilibrium reaction is
C2?H6?(g) ? C2?H4?(g)+H2?(g).
Initial | 4 | 0 | 0 |
Change | −x | x | x |
Equilibrium | 4−x | x | x |
The expression for the equilibrium constant is Kp?= (?PC2?H4?) (PH2) / PC2?H6?.
Substituting the values in the above equation, we get
Upvote 7.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
2ICl (g) ⇌ I2 (g) + Cl2 (g); Kc = 0.14
7.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?
2ICl (g) ⇌ I2 (g) + Cl2 (g); Kc = 0.14
Suppose at equilibrium, the molar concentration of both I2 and Cl2 is x mol L-1.
Kc = [I2] [ Cl2] / [ICl]2= x2 / (0.78 – 2x)2
=>x/ (0.78 – 2x) = (0.14)1/2 = 0.374
=> x= 0.167
[ICl] = (0.78 – 2 x 0.167) = (0.78 – 0.334) = 0.446 M
[I2] = 0.167 M,
[Cl2] = 0.167 M
Upvote 7.15. At 700 K, equilibrium constant for the reaction:
H2?(g) + I2 (g) ?2HI (g)
is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2?(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
7.15. At 700 K, equilibrium constant for the reaction:
H2?(g) + I2 (g) ?2HI (g)
is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2?(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
H2 (g) + I2 (g)? 2HI (g); K=64, T=700K
2HI? H2 + I2 K=1/64700K
a 0 0
a (1−α) aα/2 aα/2
0.5 x x
x2 / (0.5)
Upvote 7.14. One mole of H2O and one mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation.
H2O (g) + CO (g) ? H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.
7.14. One mole of H2O and one mole of CO are taken in a 10 litre vessel and heated to 725 K, at equilibrium point 40 percent of water (by mass) reacts with carbon monoxide according to equation.
H2O (g) + CO (g) ? H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.
Number of moles of water originally present = 1 mol
Percentage of water reacted= 40%
Number of moles of water reacted= 1 x 40/100 = 0.4 mol
Number of moles of water left= (1 – 0.4) = 0.6 mole
According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hy
Upvote 7.13. The equilibrium constant expression for a gas reaction is,
Write the balanced chemical equation corresponding to this expression.
7.13. The equilibrium constant expression for a gas reaction is,
Write the balanced chemical equation corresponding to this expression.
Balanced chemical equation for the reaction is
4 NO (g) + 6 H2O (g)? 4 NH3 (g) + 5 O2 (g)
Upvote 7.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant Kc for the reaction
N2?(g) + 3H2? (g) ? 2NH3? (g) is 1.7×102
Is this reaction at equilibrium? If not, what is the direction of net reaction?
7.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant Kc for the reaction
N2?(g) + 3H2? (g) ? 2NH3? (g) is 1.7×102
Is this reaction at equilibrium? If not, what is the direction of net reaction?
According to the given equation, concentration quotient,
Qc = [NH3]2 / [N2] [ H2]3
= (8.13/20 mol L-1)2 / (1.57 / 20 mol L-1) x (1.92 / 20 mol L-1)3
= 2.38 x 103
The equilibrium constant (Kc) for the reaction = 1.7 x 10-2
As Qc ≠ Kc, the reaction is not in a state of equilibrium.
Upvote 7.11. A sample of HI (g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?
2HI (g) ? H2 (g) + I2 (g)
7.11. A sample of HI (g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?
2HI (g) ? H2 (g) + I2 (g)
pHI = 0.04 atm, pH2 = 0.08 atm, pI2 = 0.08 atm
Kp= (pH2 x pI2) / (pHI)2 = (0.08 x 0.08) / (0.04 x 0.04)
= 4.0
Upvote