Question 1. The reaction Fe2O3(s) + 3CO(gH→2Fe(s) + 3CO2(g) is carried out separately in a closed vessel and an open vessel. In which case do you expect a higher yield of CO2(g)?
Answer:
If the reaction Fe2O3(s) + 3CO(g)→2Fe(s) + 3CO2(g) is carried out in a closed vessel, an equilibrium is established between the reactants and the products. As a result, the reaction vessel always contains a mixture of reactants and products.
On the other hand, if the reaction is carried out in an open vessel, CO2(g) formed in the reaction diffuses out from the vessel and mixes with the air and does not have the opportunity to react with Fe(s).
As a result, the reaction becomes irreversible and goes to completion. At the end of the reaction, the vessel only contains Fe(s). Since the reaction reaches completion in an open vessel, the yield of CO2(g) will be higher in this condition.
Question 2. If a reversible reaction is carried out in a closed vessel, the reactant(s) is/are never used up completely. Explain the reaction with an example.
Answer:
When a reversible reaction is carried out in a closed vessel, an equilibrium is established between the reactants and the products.
As a result, the reaction system always contains a mixture of reactants and products. In other words, a reversible reaction occurring in a closed vessel never gets completed. Consequently, the reactants are never used up.
Question 3. 2BrP2(g) ⇌ Br2(g) + 5F2(g); At constant temperature, how the increase in pressure will affect the following at equilibrium— Equilibrium constant, Position of the equilibrium, and yield of the product?
Answer:
Pressure does not affect the magnitude of the equilibrium constant.
The given reversible reaction involves a decrease in several gas molecules in the backward direction. So, increasing pressure at the equilibrium of the reaction will favor the backward reaction Consequently, the yield of the product will decrease.
Question 4. Write the conjugate bases of the following acids and give the reason: HN3, CH3OH, [Al(H2O)6]3+, NH4+, HPO42-, H2O2, OH–
Answer:
According to the Bronsted-Lowry concept, an acid is a substance that can donate protons. The species formed when an acid donates a proton is called the conjugate base of the acid. The conjugate base of an acid has one fewer H-atom than the acid.
Therefore, the conjugate bases of HN3, CH3OH, [Al(H2O)6]3+, NH4+, HPO42-, H2O2 and OH– are N–3, CH3O–, [Al(H2O)5OH]2+, NH3, PO3-4, HO–2 and O-2, respectively.
Question 5. Write the conjugate acids of the following bases and give the reason: OH–, H2PO–4, O2-, HS–, SO2-3, H2O, HCO3–, NH–2, NH3, H–, C6H5NH2, S2O82- CO32-
Answer
According to the Bronsted-Lowry concept, a base is a proton acceptor. When a base accepts a proton, the conjugate acid of the die base is formed. The conjugate acid of a base contains one more H-atom than the base.
Therefore, the conjugate acids of OH–, H2PO4 , O2-, HS–, SO2–3, H2O, HCO–3, NH–2 , NH3, H–, C6H5NH2 , S2O82- and CO32-– are H2O , H3PO4 , OH– , H2S, HSO–3, H3O+, H2CO3, NH3, NH4++, H2, C6H5NH+3, HS2O8– and HCO–3 respectively.
Question 6. To find out the equilibrium constant (K) of a reaction, it is compulsory to mention the balanced equation of the reaction—why?
Answer:
The value of the equilibrium constant (K) depends on how the balanced equation of the reaction is written, example the formation of H2 (g) from H2 (g) and I2 (g) can be written in two different ways
⇒ \(\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \text { (2) } \frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \rightleftharpoons \mathrm{HI}(g)\)
Both 1 and 2 express the same reaction but the coefficients of reactants and products are different. As a result, the value of the equilibrium constant (K) for the reaction will not be the same in the above two cases.
Question 7. Find Out Kp/Kc for the solutions \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{CO}_2(g)\)
Answer:
We know that \(K_p=K_c(R T)^{\Delta n}.\)
For the given reaction \(\Delta n=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}\)
Thus \(K_p=K_c(R T)^{-\frac{1}{2}}\)
⇒ \(\frac{K_p}{K_c}=\frac{1}{\sqrt{R T}}\)
Question 8. By what factor will the concentration of H3O+ ions in an aqueous solution be increased or decreased if its pH is increased by one unit?
Answer:
If the pH of an aqueous solution be x, then [H3O+] in the solution = 10-pH
= 10-x mol.L-1
Increasing the pH of this solution by one unit makes
pH = 1 + x.So, (H3O+) In the solution
= 10-pH = 10-(1+x)
= \(\frac{10^{-x}}{10} \mathrm{~mol} \cdot \mathrm{L}^{-1}\)
Question 9. Why is the ionic product of water at 50 greater than that at 25°C?
Answer:
Ionisation of water \(\left[2 \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\right]\)
Is an endothermic process. The equilibrium constant referring to the ionization equilibrium of water is called the ionic product of water (Kw). As the process is endothermic, Kw increases with the temperature rise. Thus, Kw of water at 50 °C is greater than that at 25 °C.
Question 10. The pH of solution A is twice that of solution B. If the concentrations of H3O+ ions in A and B are x (M) and y (M), respectively, then what is the relation between x and y?
Answer:
Given:
pH of the solution A = 2 × pH of the solution B,
⇒ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_A=x \mathrm{~mol} \cdot \mathrm{L}^{-1} \text { and }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_B=y \mathrm{~mol} \cdot \mathrm{L}^{-1} \text {. }\)
∴ \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_A=2 \times-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]_B\)
Or, \(x=y^2 \text { or, } y=\sqrt{x}\)