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11 months ago

Chemistry Equilibrium

7.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
0 Follower 3 Views

V
Vishal Baghel

Contributor-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

11 months ago

Chemistry Equilibrium

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?  
0 Follower 13 Views

V
Vishal Baghel

Contributor-Level 10

For reverse reaction, KC (reverse) = 1/ KC = 1 / 6.3 x 1014 = 1.59 x 10-15

New answer posted

11 months ago

Chemistry Equilibrium

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       
0 Follower 13 Views

V
Vishal Baghel

Contributor-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

11 months ago

Chemistry Equilibrium

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+

...more
0 Follower 19 Views

V
Vishal Baghel

Contributor-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

11 months ago

Chemistry 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.                      
0 Follower 2 Views

V
Vishal Baghel

Contributor-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

11 months ago

Chemistry Equilibrium

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?
0 Follower 1 View

V
Vishal Baghel

Contributor-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

...more

New answer posted

11 months ago

Chemistry States of Matter

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?
0 Follower 1 View

P
Payal Gupta

Contributor-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 =

...more

New answer posted

11 months ago

Chemistry States of Matter

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?
0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

5.6. The chemical equation for the reaction is
                                    2 Al + 2 NaOH + H2O? 2 NaAlO2 + 3H2 
i.e. 2 moles of Al give 3 moles of H2 gas.

Moles of aluminium = 270.15g = 5.56*10?3 moles

Moles of H2 produced= 23*5.56*10?3

           = 8.33*10?3 moles

           Volume of H2 = nRT / P

        &nbs

...more

New answer posted

11 months ago

Chemistry States of Matter Marksheet

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.
0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

5.5. Suppose molecular masses of A and B are MA and MB respectively. Then their number of moles will be

nA= 1/MA                                nB = 2/MB

Given: PA = 2 bar        and PA + PB = 3 bar

          => PB = 1bar

Since, PV = nRT

PA= nART and PBV= nBRT

Therefore, (PA / PB) = (nA / nB)= (MB / 2MA)

=> MB / MA = 2 x PA / PB = 2 x 2 /1 = 4

        => MB = 4 MA

New answer posted

11 months ago

Chemistry States of Matter

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.
0 Follower 1 View

P
Payal Gupta

Contributor-Level 10

Answer: According to ideal gas equation
                        PV = nRT

Or                        P= nRT/V

Replacing n by m/M, we get

                            P M1 x 2 = 28 x 5 (Molecular mass of N2 = 28 g/mol)
or           

...more

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