Ncert Solutions Chemistry Class 11th
Get insights from 2k questions on Ncert Solutions Chemistry Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Chemistry Class 11th
- 2k
Questions
- 0
Discussions
- 1
Active Users
- 29
Followers
Have a question about Ncert Solutions Chemistry Class 11th? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
7.6. For the following equilibrium, K =6.3 x 1014 at 1000 K.
NO (g) + O3 —–> NO2 (g) + O2(g).
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?
7.6. For the following equilibrium, K =6.3 x 1014 at 1000 K.
NO (g) + O3 —–> NO2 (g) + O2(g).
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?
For reverse reaction, KC (reverse) = 1/ KC = 1 / 6.3 x 1014 = 1.59 x 10-15
Upvote 
7.5. Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K
(ii) CaCO3 (s) ⇌ CaO (s) + CO2(g); Kp= 167 at 1073 K
7.5. Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K
(ii) CaCO3 (s) ⇌ CaO (s) + CO2(g); Kp= 167 at 1073 K
Using the equation Kp = Kc (RT)ng
(i) From the given equation, Δng = 3 – 2 = 1,
R = 0.0821 L atm K-1 mol-1
T = 500 K, Kp= 1.8 × 10–2
Thus, Kc = Kp / (RT)ng= (1.8 x 10-2) / (0.0821 L atm K-1 mol-1 x 500 K)
4.4 x 10-4 mol L-1
(ii) From the given equation, Δng =1,
R = 0.0821 L atm K-1 mol-1
T =
Upvote 7.4. Write the expression for the equilibrium constant, Kc for each of the following reactions
(i) 2NOCl (g) ?? 2NO (g) + Cl2 (g)
(ii) 2Cu(NO3)2 (s) ?? 2CuO (s) + 4NO2 (g) + O2 (g)
(iii) CH3COOC2H5 (aq) + H2O (l) ?? CH3COOH (aq) + C2H5OH (aq)
(iv) Fe3+ (aq) + 3OH– (aq) ?? Fe(OH)3(s)
(v) I2 (s) + 5F2?2IF5
7.4. Write the expression for the equilibrium constant, Kc for each of the following reactions
(i) 2NOCl (g) ?? 2NO (g) + Cl2 (g)
(ii) 2Cu(NO3)2 (s) ?? 2CuO (s) + 4NO2 (g) + O2 (g)
(iii) CH3COOC2H5 (aq) + H2O (l) ?? CH3COOH (aq) + C2H5OH (aq)
(iv) Fe3+ (aq) + 3OH– (aq) ?? Fe(OH)3(s)
(v) I2 (s) + 5F2?2IF5
(i) The expression for the equilibrium constant is Kc= [NO]2 [Cl2] / [NOCl]2.
(ii) The expression for the equilibrium constant is Kc= [NO2]4 [O2] / [ (2Cu (NO3)2]2 = [NO2]4 [O2].
(iii) The expression for the equilibrium constant is Kc= [CH3COOH] [C2H5OH] / [CH3COOC2H5].
(iv)&nb
Upvote 7.3. At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms
I2 (g) ? 2I (g)
Calculate Kp for the equilibrium.
7.3. At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms
I2 (g) ? 2I (g)
Calculate Kp for the equilibrium.
According to the given condition,
Total pressure of equilibrium mixture = 105 Pa
Partial pressure of iodine atoms (I), PI = 40 % of 105 Pa
= 0.4 x 105 Pa
Partial pressure of iodine molecules (I2), PI2 = 60 % of 105 Pa
= 0.6 x 105 Pa
Kp for the equilibrium = (PI)2/ PI2 = (0.4 x 105 Pa)2 / (0.6 x 105 Pa)
=
Upvote 7.1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on the vapour pressure?
(b) How do the rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
7.1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on the vapour pressure?
(b) How do the rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
(a) On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours is now distributed over a larger space.
(b) On increasing the volume of the container, the rate of evaporation will increase initially because now more space is available. S
Upvote 5.7. What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9 dm3 flask at 27 °C?
5.7. What will be the pressure exerted by a mixture of 3.2g of methane and 4.4g of carbon dioxide contained in a 9 dm3 flask at 27 °C?
5.7. Applying PV = nRT
We get P = nRT / V
Given: nCH4 = 3.2 / 16 mol = 0.2 mol
&
Upvote 5.6. The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminium reacts?
5.6. The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15g of aluminium reacts?
5.6. The chemical equation for the reaction is
2 Al + 2 NaOH + H2O? 2 NaAlO2 + 3
Upvote 5.5. Pressure of l g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.
5.5. Pressure of l g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.
5.5. Suppose molecular masses of A and B are MA and MB respectively. Then their number of moles will be
nA= 1/MA
Upvote 5.3. Using the equation of state PV= nRT, show that at a given temperature, density of a gas is proportional to the gas pressure P.
5.3. Using the equation of state PV= nRT, show that at a given temperature, density of a gas is proportional to the gas pressure P.
Answer: According to ideal gas equation
PV = nRT
Or &nbs
Upvote