Ncert Solutions Chemistry Class 11th
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New answer posted
6 months agoContributor-Level 10
Kp = Kc (RT)? ng
=> Kc = Kp (RT)-? ng
Putting the values of Kp= 2.0x1010 bar-1, R= 0.083 L bar K-1 mol-1, T = 450 K, and? ng = 2-3= -1
=> Kc = (2.0 x 1010 bar-1) x [ (0.083 L bar K-1 mol-1) x (450 K)]- (-1)
= 7.47 x 1011 mol-1 L
= 7.47 x 1011 M-1
New answer posted
6 months agoContributor-Level 10
According to the equation, 2 moles of NO (g) react with 1 mole of Br2 (g) to form 2 moles of NOBr (g). The composition of the equilibrium mixture can be calculated as follows:
No. of moles of NOBr (g) formed at equilibrium = 0.0518 mol
No. of moles of NO (g) taking part in reaction = 0.0518 mol
No. of moles of NO (g) left at equilibrium = 0.087 – 0.0518 = 0.0352 mol
No. of moles of Br2 (g) taking part in reaction = 1/2 x 0.0518 = 0.0259 mol
No. of moles of Br2 (g) left at equilibrium = 0.0437 – 0.0259 = 0.0178 mol
The initial molar concentration and equilibrium molar concentration of different spe
New answer posted
6 months agoContributor-Level 10
Let x moles of N2 (g) take part in the reaction. According to the equation, x/2 moles of O2 (g) will react to form x moles of N2O (g). The molar concentration per litre of different species before the reaction and at the equilibrium point is:
| [N2? ] | [O2] | [N2? O] |
Initial concentration | 0.482? / 10 | 0.933/ 10 | 0 |
Conc. at equilibrium | (0.482 – x)? / 10 | (0.933 – x/2) / 10 | x/ 10 |
The value of equilibrium constant (2.0 x 10-37) is extremely small. This means that only small amounts of reactants have reacted. Therefore, value of x is extremely small and can be omitted as far as the reactants are concerned.
Applying Law of chemical equilibrium, Kc = [N2O]2 / [N2]2 [O2]
= Kc = 2.0 x 10-37
= (x/10)2 / [ (0.482/ 10)2 x (0.933 / 10)2]
= 0.01 x2 / (2.1676 x 10-4)
x2= 4
New answer posted
6 months agoContributor-Level 10
This is because molar concentration of a pure solid or liquid is independent of the amount present.
Since density as well as molar mass of pure liquid or solid is fixed; their molar concentrations are constant.
The concentration of solid or liquid is = = =
At constant temperature, the density and molar mass of pure solid and liquid are constant.
Due to this, their molar concentrations are constant and are not included in the equilibrium constant.
New answer posted
6 months agoContributor-Level 10
For reverse reaction, KC (reverse) = 1/ KC = 1 / 6.3 x 1014 = 1.59 x 10-15
New answer posted
6 months agoContributor-Level 10
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 = 1073 K, Kp= 167 atm
Thus, Kc = Kp / (RT)ng= (167 atm) / (0.0821 L atm K-1 mol-1 x 1073 K)
= 1.9 mol L-1
New answer posted
6 months agoContributor-Level 10
(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) The expression for the equilibrium constant is Kc= 1 / [Fe3+] [OH]3.
(v) The expression for the equilibrium constant is Kc= [IF5]2 / [F2]5.
New answer posted
6 months agoContributor-Level 10
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)
= 2.67 x 104 Pa
New answer posted
6 months agoContributor-Level 10
(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. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.
(c) Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the tempera
New answer posted
6 months agoContributor-Level 10
5.7. Applying PV = nRT
We get P = nRT / V
Given: nCH4 = 3.2 / 16 mol = 0.2 mol
nCO2 = 4.4 /44 mol = 0.1 mol
So,
PCH4 = (0.2 mol) (0.0821 dm3atm K-1 mol-1) (300 K) / (9 dm3)
= 0.55 atm
PCO2= (0.1 mol) (0.0821 dm3atm K-1 mol-1) (300 K) / (9 dm3)
= 0.27 atm
Ptotal =
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