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Q. No.Consider the reactions:
(i)
\(\mathrm{{CO}(g)+{H}_{2} O(g) \rightleftharpoons {CO}_{2}(g)+{H}_{2}(g) ; K_{1}}\)
(ii)
\(\mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g); K_{2}}\)
(iii)
\(\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g) ; K_{3}}\)
Correct relationship among the equilibrium constants is:
1. \(\mathrm{K_{3}=K_{1} / K_{2} }\)
2. \(\mathrm{K_{3}=K_{1}^{2} / K_{2}^{2} }\)
3. \(\mathrm{ K_{3}=K_{1} \cdot K_{2}}\)
4. \(\mathrm{K_{3}=K_{1} \cdot \sqrt{K_{2}} }\)
Subtopic: Introduction To Equilibrium |
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Level 1: 80%+
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Given:
Kb = 1.6 × 10⁻⁸
1. 0.01%
2. 0.04%
3. 4%
4. 10%
Subtopic: Ionisation Constant of Acid, Base & Salt |
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Level 1: 80%+
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The relationship between molar solubility(s) and solubility product (Ksp) for Pb3(PO4)2 is:
| 1. | \(s=\left(\frac{K_{s p}}{106}\right)^{1 / 4}~ \) | 2. | \(s=\left(\frac{K_{s p}}{108}\right)^{1 / 5} \) |
| 3. | \(s=\left(\frac{K_{s p}}{81}\right)^{1 / 5} \) | 4. | \(s=\left(\frac{K_{s p}}{48}\right)^{1 / 5} \) |
Subtopic: Solubility Product |
92%
Level 1: 80%+
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What is the solubility of BaSO4 in 10-3 M H2SO 4 solution ?
(Given: Ksp for BaSO4 = 1.1 × 10-10)
(Given: Ksp for BaSO4 = 1.1 × 10-10)
| 1. | 1.1 × 10–13 M | 2. | 1.1 × 10–7 M |
| 3. | 5.5 × 10–7 M | 4. | 5.5 × 10–8 M |
Subtopic: Solubility Product |
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Level 1: 80%+
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The correct statement regarding the equilibrium constant(K) of a specific reaction is:
| 1. | It may be changed by the addition of a catalyst. |
| 2. | It increases if the concentration of one of the products increases. |
| 3. | It changes with changes in temperature. |
| 4. | It increases if the concentration of one of the reactants is increased. |
Subtopic: Introduction To Equilibrium |
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Level 1: 80%+
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In some solutions, the concentration of H3O+ remains constant even when small amounts of strong acid or strong base are added to them. These solutions are known as:
1. Ideal solution
2. Colloidal solution
3. True solution
4. Buffer solution
1. Ideal solution
2. Colloidal solution
3. True solution
4. Buffer solution
Subtopic: Buffer |
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Find the equilibrium constant for the reaction
2. (Kc)ⁿ
3. Kc/n
4. nKc
naA + nbB ⇌ ncC + ndD
if the equilibrium constant for
aA + bB ⇌ cC + dD
is Kc.
1. n√Kc2. (Kc)ⁿ
3. Kc/n
4. nKc
Subtopic: Introduction To Equilibrium |
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The value of Kp for the following reaction at 100 K will be:
\(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g})\)
(Equilibrium constant, Kc at 100 K is \(3 \times 10^{-4} \) and R is 8.314 J/mol K)
1. 0.25
2. 0.025
3. 2.5
4. 25
\(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g})\)
(Equilibrium constant, Kc at 100 K is \(3 \times 10^{-4} \) and R is 8.314 J/mol K)
1. 0.25
2. 0.025
3. 2.5
4. 25
Subtopic: Kp, Kc & Factors Affecting them |
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The given reaction:
\(NaOH\ + \ HCl\ \rightarrow \ NaCl \ +\ H_2O\) is an example of:
1. Intramolecular redox reaction
2. Disproportionation reaction
3. Acid-base reaction
4. Decomposition reaction
\(NaOH\ + \ HCl\ \rightarrow \ NaCl \ +\ H_2O\) is an example of:
1. Intramolecular redox reaction
2. Disproportionation reaction
3. Acid-base reaction
4. Decomposition reaction
Subtopic: Acids & Bases - Definitions & Classification |
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Level 1: 80%+
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A sample of \(AB_2(g) \) was introduced into an evacuated vessel. After equilibrium is attained concentration of \(AB_2(g) \) is found to be 0.5 mol L-1. If the value of KC is 0.5, then the concentration of A(g) at equilibrium for the reaction \(AB_2(g) \rightleftharpoons A(g) + B_2(g) \) is:
1. 0.1 M
2. 0.2 M
3. 0.5 M
4. 1 M
1. 0.1 M
2. 0.2 M
3. 0.5 M
4. 1 M
Subtopic: Introduction To Equilibrium |
81%
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