At a certain temperature and pressure of 105 Pa, iodine vapour contains 40% by volume of I atoms. The Kp for the equilibrium of the reaction would be-
1. 2.67 104 Pa
2. 1.00 105 Pa
3. 3.63 104 Pa
4. 2.18 105 Pa
Given the reaction 2HI (g) H2 (g) + I2 (g)
A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI(g) is 0.04 atm. The for the given equilibrium would be:
1. | 2.0 | 2. | 3.5 |
3. | 4.0 | 4. | 2.6 |
A mixture of 1.57 mol of N2, 1.92 mol of H2, and 8.13 mol of NH3 is introduced into a 20 L vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) is 1.7 × 102.
The direction of the net reaction is:
1. Reaction is at equilibrium.
2. Reaction will proceed in forwarding direction.
3. Reaction will proceed in the backward direction.
4. Data is not sufficient.
Given the reaction:
2BrCl (g) Br2 (g) + Cl2 (g); Kc= 32 at 500 K. If the initial concentration of BrCl is 3.3 × 10-3 mol L–1, the molar concentration of BrCl in the mixture at equilibrium would be:
1. 3.0 10-2 molL-1
2. 2.0 10-4 molL-1
3. 2.5 10-6 molL-1
4. 3.0 10-4 molL-1
For the reaction, NO(g) + 1/2O2 (g) ⇌ NO2(g)
∆fG° (NO2) = 52.0 kJ/mol , ∆fG° (NO) = 87.0 kJ/mol and
∆fG° (O2) = 0 kJ/mol. The equilibrium constant for the formation of NO2 from
NO and O2 at 298K would be
1. 2.36 104
2. 3.10 107
3. 1.36 10
4. 2.18 10
The equilibrium constant for the following reaction is 1.6 ×105 at 1024K
H2(g) + Br2(g) 2HBr(g)
If HBr at pressure 10.0 bar is introduced into a sealed container at 1024 K, the equilibrium pressure of HBr will be :
1. 11.20 bar
2. 5.56 bar
3. 7.30 bar
4. 9.95 bar
For the reaction, 2NOCl (g) 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 atm at 500 K.
The value of Kc for above mentioned reaction would be:
For the following equilibrium, Kc= 6.3 × 1014 at 1000 K
The value of Kc for the reverse reaction is:
\(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5}{[\mathrm{NO}]^4\left[\mathrm{H}_2 \mathrm{O}]^6\right.}\)
The balanced chemical equation corresponding to the above-mentioned expression is:
1. | \(4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons 4 \mathrm{NH}_{3(\mathrm{g})}+5 \mathrm{O}_{2(\mathrm{g})} \) |
2. | \(4 \mathrm{NH}_3(\mathrm{g})+5 \mathrm{O}_{2(\mathrm{g})} \rightleftharpoons 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}\) |
3. | \(\ 2 \mathrm{NO}_{(\mathrm{g})}+3 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons 4 \mathrm{NH}_{3(\mathrm{g})}+3 \mathrm{O}_{2(\mathrm{g})}\) |
4. | \(\ \mathrm{NH}_{3(\mathrm{g})}+3 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons 2 \mathrm{NO}_{(\mathrm{g})}+3 \mathrm{O}_{2(\mathrm{g})}\) |
One mole of H2O and
one mole of CO are taken in a 10 L vessel and heated to
725 K. At equilibrium, 40% of water(by mass) reacts with CO according to the equation,
H2O (g) + CO (g) H2 (g) + CO2 (g)
The equilibrium constant for the above-mentioned reaction would be:
1. | 0.66 | 2. | 0.35 |
3. | 0.44 | 4. | 0.82 |