Question

1. Consider this reversible reaction. All substances are gases and are in an insulated closed chamber with a constant volume and are at equilibrium. 2H2O + 2Cl2 4HCl + O2 ΔH = +113 kJ What will be the effect of increasing the temperature?

(A) More Cl2 will be produced. (B) More HCl will be produced. (C) The volume of each of the gases will increase. (D) More water will be formed. (E) There will be no change.

2. In the following voltaic cell, to which electrode do the electrons flow?

2Al³⁺_(aq) + 3Ca_(s) à 2Al_(s) + 3Ca²⁺_(aq)

A) Al³⁺              B) Ca                           C) Al                           D) Ca²⁺

3.¹⁷⁹Ta decays by electron capture. What is the product of this reaction?

A)   ¹⁷⁹W B)  ¹⁸³Re            C)  ¹⁸⁰Ta D)  ¹⁷⁹Hf E)  ¹⁷⁵Lu

4. The reaction SO₂Cl₂ = SO₂ + Cl₂ is first order in SO₂Cl₂. If the concentration of SO₂Cl₂ after 198 s is 1.47 x 10^-2 M and the rate constant is 2.34 x 10^-2 s^-1, what was the initial concentration of SO₂Cl₂?

A) 1.40 M          B) 1.53 M              C) 4.65 M                D)   6.33 x 10² M            E) 1.47 x 10² M

5. What is the K_sp for Mg(OH)₂ when the solubility is 1.4 x 10^-4 M?

A) 2.7 x 10^-12      B) 1.1 x 10^-11                   C) 2.0 x 10^-8             D) 3.9 x 10^-8           E) 1.4 x 10^-8

Answer 1

Ans. #1. Note that the reaction is endothermic as dH is positive (dH = + 113 kJ). An endothermic reaction requires heat to proceed in forward direction.

Given- “Temperature is increased”. Increasing temperature is a means of supplying heat to the reaction mixture. So, being endothermic reaction, the reaction will proceed in forward direction favoring formation of HCl when temperature is increased.

Thus, correct option is- B. More HCl will be produced.

Option A and D are incorrect because they indicate the reaction to proceed in backward direction.

#2. Note: Al³⁺ is reduced to Al⁰ by accepting electrons.

Ca⁰ is being oxidized to Ca²⁺ by releasing electrons.

So, electrons flow from Ca (donor) to Al³⁺.

Note that electrodes (metal rods) are immersed in their respective salt solution.

That is, Al(s) is the electrode and Al³⁺ is a component of its salt solution. Similarly, Ca(s) is also the electrode.

Therefore, electrons flow from Ca(s) to Al(s) electrode.

Correct option – C. Al

#3. Electron capture: In this process, an electron in drawn into the nucleus from atom’s inner orbital. The electron then combines with a proton to form a neutron, that is, a proton is converted into a neutron- thus, atomic number is reduced by 1 unit while atomic mass remains the same.

            ⁷³Ta₁₇₉ + 1e ----------> ⁷²Hf₁₇₉

Thus, Ta (atomic number = 73) is converted into Hf179 (atomic number =72).

So, correct option is – D. Hf179.

Note: Mass number remains the same, so options B, C and E are incorrect because they all has a mass number different from parent nucleus.

W has atomic number 74, 1 unit greater than Ta. Since atomic number must reduce by 1 unit during electron capture, this option is incorrect.

#4. First order kinetics-

ln [A]_t - ln [A]₀) = -kt                            -equation 1

where, [A]_t = Final concentration of reaction

            [A]₀ = Initial concentration of reactant

            k = rate constant

            t = time of reaction

Putting the values in equation 1-

            ln (1.47 x 10^-2) – ln [A]₀ = - (2.34 x 10^-2 s^-1) x 198 s

            or, (- 4.22) – ln [A]₀ = - 4.6332

            or, 4.22 + ln [A]₀ = 4.6332

            or, ln [A]₀ = 4.6332 - 4.22 = 0.4132

            or, 2.303 log [A]₀ = 0.4132

            or, log [A]₀ = 0.4132 / 2.303 = 0.1794

            or, [A]₀ = antilog 0.1794 = 1.511

Therefore, initial concentration = 1.511 M

Correct option – B. 1.53 M (nearest option)

#5. Balanced reaction:         Mg(OH)₂ --------------> Mg²⁺(aq) + 2 OH^-(aq)

1 mol Mg(OH)₂ yields 1 mol Mg²⁺ and 2 mol OH^-.

Given,

            Solubility of Mg(OH)₂ = 1.4 x 10^-4 M. That is, there are 1.4 x 10^-4 moles/L Mg(OH)₂

Solvated in the solution.

So,

            [Mg²⁺] = 1.4 x 10^-4 M           ; [1 mol Mg(OH)₂ gives 1 mol Mg²⁺]

            [OH^-] = 2 x [Mg²⁺] = 2 x 1.4 x 10^-4 M = 2.8 x 10^-4 M

Now, Ksp = [Mg²⁺] [OH^-]² = (1.4 x 10^-4) x (2.8 x 10^-4)² = 1.0 x 10^-11

So, correct option is- B. 1.0 x 10^-11