Q. No.The pH of pure water is 7.0 at 25°C. When the temperature of pure water is increased to 80°C, what happens to its pH value?
1. The pH decreases
2. The pH remains the same (still 7.0)
3. The pH first increases and then decreases
4. The pH increases
Subtopic: Introduction To Equilibrium | pH calculation |
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Level 2: 60%+
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Consider the given reaction:
\(\mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{~g})\)
If initially only \(\mathrm{H_2}\) and \(\mathrm{I_2}\) are present, which graph correctly predicts the attainment of equilibrium?
\(\mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{~g})\)
If initially only \(\mathrm{H_2}\) and \(\mathrm{I_2}\) are present, which graph correctly predicts the attainment of equilibrium?
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2. |  |
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4. |  |
Subtopic: Introduction To Equilibrium |
82%
Level 1: 80%+
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Consider the following reaction,
\(H_2O(g) ⇋ H_2(g) +\frac12O_2(g)~ \)
If Keq = 2 × 10−3 at 2300 K and the initial pressure of H2O(g) is 1 atm, then the degree of dissociation of the above reaction will be x × 10−2, and the value of x is:
1. 3
2. 4
3. 2
4. 7
\(H_2O(g) ⇋ H_2(g) +\frac12O_2(g)~ \)
If Keq = 2 × 10−3 at 2300 K and the initial pressure of H2O(g) is 1 atm, then the degree of dissociation of the above reaction will be x × 10−2, and the value of x is:
1. 3
2. 4
3. 2
4. 7
Subtopic: Introduction To Equilibrium |
60%
Level 2: 60%+
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Consider the following reactions:
X(g) ⇌ 2Y(g) KP1 ----(i)
A(g) ⇌ B(g) + C(g) KP2 ----(ii)
If the degree of dissociation is the same for both reactions. The ratio of total pressure P1 & P2 respectively is:
1. \(\frac{K_{P_1}}{K_{P_2}}\)
2. \(\frac{4 K_{P_1}}{K_{P_2}}\)
3. \(\frac{K_{P_1}}{4 K_{P_2}}\)
4. \(\frac{K_{P_1}}{2 K_{P_2}}\)
X(g) ⇌ 2Y(g) KP1 ----(i)
A(g) ⇌ B(g) + C(g) KP2 ----(ii)
If the degree of dissociation is the same for both reactions. The ratio of total pressure P1 & P2 respectively is:
1. \(\frac{K_{P_1}}{K_{P_2}}\)
2. \(\frac{4 K_{P_1}}{K_{P_2}}\)
3. \(\frac{K_{P_1}}{4 K_{P_2}}\)
4. \(\frac{K_{P_1}}{2 K_{P_2}}\)
Subtopic: Introduction To Equilibrium |
67%
Level 2: 60%+
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Initially 2 moles of \(NOCl \) taken in 1 L of closed container and it dissociates into NO and Cl2 gas:
The reaction is as follows:
\(2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g)\)
Calculate the equilibrium constant \(K_c\) given that 0.4 moles of NO is obtained at equilibrium:
The reaction is as follows:
\(2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g)\)
Calculate the equilibrium constant \(K_c\) given that 0.4 moles of NO is obtained at equilibrium:
| 1. | 142 × 10-4 | 2. | 142 × 10-6 |
| 3. | 125 × 10-4 | 4. | 130 × 104 |
Subtopic: Introduction To Equilibrium |
82%
Level 1: 80%+
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For an equilibrium reaction :
\(A \rightleftharpoons B+\frac{1}{2} C \)
The correct relation between the degree of dissociation\((\alpha)\), equilibrium pressure(p), and equilibrium constant(\(K_p\)) is :
1. \(K_p =\frac{\alpha^{1 / 2} 2 p^{1 / 2}}{(2+\alpha)^{1 / 2}} \)
2. \(K_p =\frac{\alpha^{3 / 2} p^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}\)
3. \(K_p =\frac{\alpha^{1 / 2} 2 p^{1 / 2}}{(2+\alpha)^{3 / 2}}\)
4. \(K_p =\frac{\alpha^{1 / 2} p^{3 / 2}}{(2+\alpha)^{3 / 2}}\)
\(A \rightleftharpoons B+\frac{1}{2} C \)
The correct relation between the degree of dissociation\((\alpha)\), equilibrium pressure(p), and equilibrium constant(\(K_p\)) is :
1. \(K_p =\frac{\alpha^{1 / 2} 2 p^{1 / 2}}{(2+\alpha)^{1 / 2}} \)
2. \(K_p =\frac{\alpha^{3 / 2} p^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}\)
3. \(K_p =\frac{\alpha^{1 / 2} 2 p^{1 / 2}}{(2+\alpha)^{3 / 2}}\)
4. \(K_p =\frac{\alpha^{1 / 2} p^{3 / 2}}{(2+\alpha)^{3 / 2}}\)
Subtopic: Introduction To Equilibrium | Kp, Kc & Factors Affecting them |
Level 4: Below 35%
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At 1990 K and 1 atm pressure, there are an equal number of Cl2 molecules and Cl atoms in the reaction mixture.
The value of KP for the reaction under the above conditions is x × 10–1. The value of x is:
1. 4
2. 8
3. 5
4. 10
Subtopic: Introduction To Equilibrium |
58%
Level 3: 35%-60%
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For the reaction A(g)(B)(g), the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of rG for the reaction at 300 K and 1 atm in J mol–1 is – xR, where x is:
(R = 8.31 J mol–1 K–1 and ln 10 = 2.3)
1. 1400
2. 1380
3. 1360
4. 1340
Subtopic: Introduction To Equilibrium |
84%
Level 1: 80%+
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The equilibrium constant \(K_C\) at \(298~K\) for the reaction \(A+B \rightleftharpoons C+D\) is \(100.\)
Starting with an equimolar solution with concentrations of A, B, C and D all equal to \(1~M,\) if the equilibrium concentration of D comes out to be \(x \times10^{-2}M\), then find the value of \(x\):
1. 181
2. 190
3. 188
4. 192
Starting with an equimolar solution with concentrations of A, B, C and D all equal to \(1~M,\) if the equilibrium concentration of D comes out to be \(x \times10^{-2}M\), then find the value of \(x\):
1. 181
2. 190
3. 188
4. 192
Subtopic: Introduction To Equilibrium |
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\(100 ~g\) of propane is completely reacted with \(1000~ g\) of oxygen. The mole fraction of carbon dioxide in the resulting mixture is \(x \times 10^{–2}\) . The value of \(x\) is :
1. 18.52
2. 19.02
3. 19.82
4. 20.05
Subtopic: Introduction To Equilibrium |
75%
Level 2: 60%+
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