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Equilibrium

Equilibrium relates to CBSE/Class 11/Science/Chemistry

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Equilibrium Questions

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Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

To predict the effect of temperature on the equilibrium constant (KK) of the given reaction, we can use the relationship between the standard Gibbs free energy change (ΔG∘ΔG∘) and the equilibrium constant: ΔG∘=−RTln⁡KΔG∘=−RTlnK where: ΔG∘ΔG∘ is the... read more

To predict the effect of temperature on the equilibrium constant (KK) of the given reaction, we can use the relationship between the standard Gibbs free energy change (ΔG∘ΔG) and the equilibrium constant:

ΔG∘=−RTln⁡KΔG=−RTlnK

where:

  • ΔG∘ΔG is the standard Gibbs free energy change,
  • RR is the gas constant (8.314 J/mol·K),
  • TT is the temperature in Kelvin,
  • KK is the equilibrium constant.

At equilibrium, ΔG∘ΔG is zero. Therefore, we can rearrange the equation to solve for KK:

K=e−ΔG∘RTK=eRTΔG

Given that the reaction is:

CaCO3(s)→CaO(s)+CO2(g)CaCO3(s)→CaO(s)+CO2(g)

with the given standard enthalpy changes (ΔH∘ΔH):

ΔHCaCO3∘=−1206.9 kJ/molΔHCaCO3=−1206.9kJ/mol ΔHCaO∘=−635.1 kJ/molΔHCaO=−635.1kJ/mol ΔHCO2∘=−393.5 kJ/molΔHCO2=−393.5kJ/mol

we can use these values to calculate the change in standard Gibbs free energy (ΔG∘ΔG) using the equation:

ΔG∘=ΔHproducts∘−ΔHreactants∘ΔGHproducts−ΔHreactants

ΔG∘=(−635.1+(−393.5))−(−1206.9) kJ/molΔG=(−635.1+(−393.5))−(−1206.9)kJ/mol ΔG∘=118.3 kJ/molΔG=118.3kJ/mol

Now, we can calculate KK at two different temperatures and compare their values:

  1. At T1=298 KT1=298K: K1=e−118.3×1038.314×298K1=e8.314×298118.3×103

  2. At T2=350 KT2=350K: K2=e−118.3×1038.314×350K2=e8.314×350118.3×103

To determine the effect of temperature on KK, we can compare the values of K1K1 and K2K2. If K2>K1K2>K1, then increasing the temperature increases the equilibrium constant KK, indicating that the reaction shifts towards the products at higher temperatures. Conversely, if K2<K1K2<K1, then increasing the temperature decreases the equilibrium constant KK, indicating that the reaction shifts towards the reactants at higher temperatures.

By calculating K1K1 and K2K2 using the above equations, we can determine the effect of temperature on the equilibrium constant KK of the given reaction.

 
 
 
 
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Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

In the given reaction: NH3+BF3→H3N:BF3NH3+BF3→H3N:BF3 Ammonia (NH3NH3) acts as the base, while boron trifluoride (BF3BF3) acts as the acid. This reaction involves the formation of a coordinate covalent bond between the lone pair of electrons on the nitrogen atom in ammonia and the empty... read more

In the given reaction:

NH3+BF3→H3N:BF3NH3+BF3→H3N:BF3

Ammonia (NH3NH3) acts as the base, while boron trifluoride (BF3BF3) acts as the acid. This reaction involves the formation of a coordinate covalent bond between the lone pair of electrons on the nitrogen atom in ammonia and the empty orbital on the boron atom in boron trifluoride.

This reaction is best explained by Lewis acid-base theory. According to this theory, an acid is defined as a substance that can accept an electron pair, while a base is a substance that can donate an electron pair. In the given reaction, boron trifluoride acts as the Lewis acid by accepting the lone pair of electrons from ammonia, which acts as the Lewis base.

The hybridization of boron (BB) and nitrogen (NN) in the reactants can be determined by considering the electron configuration and bonding in the molecules:

  1. In boron trifluoride (BF3BF3), boron has an electron configuration of 1s22s22p11s22s22p1. Boron forms three bonds with fluorine atoms, resulting in sp^2 hybridization for boron. The three hybridized orbitals on boron overlap with the p orbitals of the three fluorine atoms to form three sigma bonds.

  2. In ammonia (NH3NH3), nitrogen has an electron configuration of 1s22s22p31s22s22p3. Nitrogen forms three bonds with hydrogen atoms, resulting in sp^3 hybridization for nitrogen. The three hybridized orbitals on nitrogen overlap with the s orbitals of the three hydrogen atoms to form three sigma bonds.

Therefore, in the reactants, the hybridization of boron is sp^2 and the hybridization of nitrogen is sp^3.

 
 
 
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Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

To calculate the volume of water required to dissolve 0.1 g0.1g of lead (II) chloride (PbCl2PbCl2) to get a saturated solution, we first need to determine the number of moles of PbCl2PbCl2 dissolved. Then, we can use the solubility product constant (KspKsp) to find the concentration of Pb2+Pb2+... read more

To calculate the volume of water required to dissolve 0.1 g0.1g of lead (II) chloride (PbCl2PbCl2) to get a saturated solution, we first need to determine the number of moles of PbCl2PbCl2 dissolved. Then, we can use the solubility product constant (KspKsp) to find the concentration of Pb2+Pb2+ ions in the saturated solution, which will allow us to calculate the volume of water required.

First, let's calculate the number of moles of PbCl2PbCl2 dissolved:

  1. Calculate the molar mass of PbCl2PbCl2: Molar mass of Pb=207 g/molMolar mass of Pb=207g/mol Molar mass of Cl=35.453 g/molMolar mass of Cl=35.453g/mol Molar mass of ( \text{PbCl}_2 = \text{Molar mass of Pb} + 2 \times \text{Molar mass of Cl} ] =207+2×35.453=207+70.906=277.906 g/mol=207+2×35.453=207+70.906=277.906g/mol

  2. Calculate the number of moles of PbCl2PbCl2: Number of moles=MassMolar mass=0.1 g277.906 g/molNumber of moles=Molar massMass=277.906g/mol0.1g Number of moles≈3.598×10−4 molNumber of moles≈3.598×10−4mol

Now, let's use the solubility product constant (KspKsp) to find the concentration of Pb2+Pb2+ ions in the saturated solution:

PbCl2→Pb2++2Cl−PbCl2→Pb2++2Cl

Given Ksp=3.2×10−8Ksp=3.2×10−8, we can set up an ice table:

PbCl2→Pb2++2Cl−InitialChange+x+2xEquilibrium3.598×10−42×3.598×10−4PbCl2InitialChangeEquilibrium→Pb2+++x3.598×10−42Cl+2x2×3.598×10−4

Substituting the equilibrium concentrations into the KspKsp expression:

Ksp=[Pb2+]×[Cl−]2Ksp=[Pb2+]×[Cl]2

3.2×10−8=(3.598×10−4)×(2×3.598×10−4)23.2×10−8=(3.598×10−4)×(2×3.598×10−4)2

Solve for [Pb2+][Pb2+]:

[Pb2+]=3.2×10−8(2×3.598×10−4)2[Pb2+]=(2×3.598×10−4)23.2×10−8

[Pb2+]≈2.222×10−2 M[Pb2+]≈2.222×10−2M

Now, we can use the concentration to calculate the volume of water required to dissolve 0.1 g0.1g of PbCl2PbCl2:

[Pb2+]=Amount of substanceVolume of solution[Pb2+]=Volume of solutionAmount of substance

Volume of solution=Amount of substance[Pb2+]Volume of solution=[Pb2+]Amount of substance

Volume of solution=3.598×10−4 mol2.222×10−2 MVolume of solution=2.222×10−2M3.598×10−4mol

Volume of solution≈0.0162 LVolume of solution≈0.0162L

Therefore, approximately 0.0162 L0.0162L of water is required to dissolve 0.1 g0.1g of lead (II) chloride to get a saturated solution.

 
 
 
 
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Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

To find the solubility of Al(OH)3Al(OH)3 in grams per liter (g/L), we first need to determine the concentration of Al3+Al3+ ions in the saturated solution. Then, we can use stoichiometry to find the concentration of Al(OH)3Al(OH)3 and finally convert it to grams per liter. Given: Solubility product... read more

To find the solubility of Al(OH)3Al(OH)3 in grams per liter (g/L), we first need to determine the concentration of Al3+Al3+ ions in the saturated solution. Then, we can use stoichiometry to find the concentration of Al(OH)3Al(OH)3 and finally convert it to grams per liter.

Given:

  • Solubility product (KspKsp) of Al(OH)3Al(OH)3 = 2.7×10−112.7×10−11
  • Atomic mass of Al (AlAl) = 27 u
  1. Calculate the concentration of Al3+Al3+ ions:

The dissolution of Al(OH)3Al(OH)3 proceeds as follows:

Al(OH)3⇌Al3++3OH−Al(OH)3⇌Al3++3OH

Let xx be the solubility of Al(OH)3Al(OH)3 in moles per liter (M). Then, the concentration of Al3+Al3+ ions will also be xx M.

Using the solubility product expression:

Ksp=[Al3+]×[OH−]3Ksp=[Al3+]×[OH]3

Given that [OH−]=3x[OH]=3x, we can substitute these values into the expression:

2.7×10−11=x×(3x)32.7×10−11=x×(3x)3

Solving for xx:

2.7×10−11=27x42.7×10−11=27x4

x4=2.7×10−1127x4=272.7×10−11

x4=1×10−12x4=1×10−12

x=1×10−124x=41×10−12

x=0.01 Mx=0.01M

  1. Calculate the molar mass of Al(OH)3Al(OH)3:

Molar mass of Al(OH)3=Molar mass of Al+3×Molar mass of O+3×Molar mass of HMolar mass of Al(OH)3=Molar mass of Al+3×Molar mass of O+3×Molar mass of H

=27+3×16+3×1=27+3×16+3×1

=27+48+3=27+48+3

=78 g/mol=78g/mol

  1. Calculate the solubility in grams per liter (g/L):

The solubility of Al(OH)3Al(OH)3 in grams per liter (g/L) is the molar mass of Al(OH)3Al(OH)3 multiplied by its molar solubility:

Solubility=Molar solubility×Molar massSolubility=Molar solubility×Molar mass

Solubility=0.01 M×78 g/molSolubility=0.01M×78g/mol

Solubility=0.78 g/LSolubility=0.78g/L

  1. Calculate the pH of the solution:

Since Al(OH)3Al(OH)3 is a sparingly soluble salt, we can assume that it dissociates completely in solution. Therefore, the concentration of Al3+Al3+ ions is equal to the solubility, which is 0.01 M0.01M.

To find the pH of the solution, we need to calculate the concentration of H+H+ ions. Since Al3+Al3+ ions are neutral, for each mole of Al3+Al3+ ions, three moles of H+H+ ions are produced.

[H+]=3×0.01 M=0.03 M[H+]=3×0.01M=0.03M

Now, we can use the formula for pH:

pH=−log⁡[H+]pH=−log[H+]

pH=−log⁡(0.03)pH=−log(0.03)

pH≈−log⁡(3×10−2)pH≈−log(3×10−2)

pH≈−(−1.5229)pH≈−(−1.5229)

pH≈1.5229pH≈1.5229

Therefore, the solubility of Al(OH)3Al(OH)3 in g/L is 0.78 g/L0.78g/L, and the pH of the solution is approximately 1.521.52.

 
 
 
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Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively. When equal volumes of two solutions are mixed, the resulting pH can be calculated using the formula for dilution: pH=12(pHA+pHB)pH=21(pHA+pHB) pH=12(6+4)=12(10)=5pH=21(6+4)=21(10)=5 ... read more

 

  • Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.

    When equal volumes of two solutions are mixed, the resulting pH can be calculated using the formula for dilution: pH=12(pHA+pHB)pH=21(pHA+pHB) pH=12(6+4)=12(10)=5pH=21(6+4)=21(10)=5

 

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