The Most Important Questions for Chemical Science Students - CHEMICAL EQUILIBRIUM!

Chemical Equilibrium - Most Important Questions! (Download PDF)

1. Calculate Kc for the reaction 2SO3(g) ↔  2SO₂(g) + O₂(g) for which

    K_p = 3.5 × 10^–23 atm at 27°C.

1. 1.42×10^-24
2. 1.42×10^-28
3. 1.42×10^-30
4. 1.42×10^-39

1. At 30°C, Kp for the dissociation reaction,

    SO₂Cl₂(g) ↔ SO2(g) + Cl₂(g) is 2.9 × 10^–2 atm. If the total pressure is 1 atm, then calculate the degree of dissociation of SO2Cl2(Assume α<<1).

1. 0.29
2. 0.17
3. 0.32
4. 0.56

1. For a homogeneous gaseous reaction,

          AB₂(g) A(g) + 2B(g) 

      The volume is 10 L and the temperature is 300 K.  Initially, the flask contains 0.4 moles of AB₂. The total pressure of the reaction mixture, after the attainment of equilibrium, is 1.2 atm. Calculate Kp.

1. 5.607×10^-3
2. 4.308×10^-3
3. 5.607×10^-4
4. 4.308×10^-44

1. Consider the homogeneous gaseous reaction H2(g) + I2(g) 2HI(g) carried out in a vessel at a temperature T. When 1 mole of H2 and 3 moles of I2 are mixed, a certain amount of HI will be formed. When 2 additional moles of H2 are introduced, the amount of HI formed becomes doubled the earlier amount. What is the value of Kp?

1. 6
2. 2
3. 4
4. 10

1. At 25°C, for the reaction Br2(l) + Cl2(g) 2BrCl(g), Kp = 2.032. At the same temperature, the vapour pressure of Br2(l) is 0.281 atm. Pure BrCl(g) was introduced into a closed container with adjustable volume. The total pressure was kept at 1 atm and the temperature at 25°C. Calculate the fraction of BrCl originally present that has been converted into Br2 and Cl2 at equilibrium assuming that the gaseous species behave ideally.

1. 0.256
2.  0.357
3.   0.687
4.   0.456

1. For the reaction,

        H₂(g) ↔ S(g) H₂S(g)

   The equilibrium constant K_p is 20.2 atm^–1 at 945°C and 9.1 atm^–1 at 1.065°C. Calculate  ΔH°.

1. -88.126 kJ/mol
2. +88.126 kJ/mol
3. -48.923 kJ/mol
4. +48.923 kJ/mol

1. The following data were obtained for the temperature dependence of the equilibrium constant of an inhibitor binding an enzyme. Determine the value of ΔG° (in kJ/mol) for this process at 25°C.

1. -23.5
2. +23.5
3. -43.5
4. +43.5

1.   Calculate the equilibrium pressure at which graphite gets converted to diamond at 25°C, given that the densities of graphite and diamond are, respectively, 2.25 and 3.51 g cm–3 and are independent of pressure. ΔG°f values for graphite and diamond are zero and 2.90 kJ mol–1, respectively.

1. 32000 atm
2. 25000 atm
3. 18000 atm
4. None of the above

1. Calculate the partial pressure of HCl gas above a sample of NH₄Cl(s) because of its decomposition according to the reaction NH₄Cl(s) NH₃(g) + HCl(g). ΔG°f values of NH₄Cl(s), NH₃(g) and HCl(g) are – 202.96 kJ mol^–1. –16.48 kJ mol^–1 and –95.30 kJ mol^–1. respectively.

1. 1.02×10^-6
2. 2.02×10^-8
3. 2.02×10^-6
4. 1.02×10^-8

1. At 450°C and 600 atm pressure, the equilibrium constant for the reaction N₂(g) + 3H₂(g) ↔ 2NH₃(g) is 4.516 × 10^–5. Calculate the degree of the conversion of N₂ and H₂ to NH₃ assuming that the system is a mixture of real gases. Fugacity coefficients are H₂ = 1.2874. 

1. 0.56
2. 0.43
3. 0.68
4. 0.24

Solutions

Solution 1:

Solution 2:

For the given equilibrium, 

             K_p = α²P/(1 – α²)

    In the present case, Kp = 2.9 × 10^–2 atm ; P = 1 atm

2.9 × 10^–2 atm = α² (1 atm)/(1 – α²).

Since K_p, in this case_, is very small, α would also be very small so that α2 ≪ 1 and can be neglected in the denominator giving Kp = α²P

Degree of dissociation at P = 1 atm, α = (Kp)^1/2 = (2.9 × 10^–2)^1/2 = 0.17

Solution 3:

Solution 4:

Solution 5:

Solution 6:

Solution 7:

Solution 8:

Solution 9:

Solution 10:

ANSWER KEYS

1. A
2. B
3. A
4. C
5. B
6. A
7. C
8. B
9. D
10. B

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