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Post a LessonAnswered on 10 Apr Learn 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∘=−RTlnKΔG∘=−RTlnK
where:
At equilibrium, ΔG∘ΔG∘ is zero. Therefore, we can rearrange the equation to solve for KK:
K=e−ΔG∘RTK=e−RTΔG∘
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∘ΔG∘=ΔHproducts∘−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:
At T1=298 KT1=298K: K1=e−118.3×1038.314×298K1=e−8.314×298118.3×103
At T2=350 KT2=350K: K2=e−118.3×1038.314×350K2=e−8.314×350118.3×103
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.
Answered on 10 Apr Learn 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 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:
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.
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.
Answered on 10 Apr Learn 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+ 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:
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
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:
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 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:
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
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
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
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.
Answered on 10 Apr Learn 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
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Answered on 10 Apr Learn Equilibrium
Sadika
pH of 0.08 mol dm–3HOCl solution is 2.85. Calculate its ionization constant.
The ionization constant (KaKa) for a weak acid (HAHA) can be calculated using the pH and initial concentration ([HA]0[HA]0) of the acid solution: Ka=[H+][OCl−][HOCl]=10−pHKa=[HOCl][H+][OCl−]=10−pH Ka=10−2.85Ka=10−2.85 Ka=1.318×10−3 mol−1Ka=1.318×10−3mol−1
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Answered on 10 Apr Learn Equilibrium
Sadika
A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Qsp) becomes greater than its solubility product. If the solubility of BaSO4BaSO4 in water is 8×10−4 mol dm−38×10−4mol dm−3, calculate its solubility in 0.01 mol dm−30.01mol dm−3 of H2SO4H2SO4.
Solubility product (KspKsp) is a constant at a given temperature. If Qsp>KspQsp>Ksp, precipitation occurs. Qsp=[Ba2+][SO42−]=(8×10−4)(8×10−4)=6.4×10−7 mol2dm−6Qsp=[Ba2+][SO42−]=(8×10−4)(8×10−4)=6.4×10−7mol2dm−6 Since Qsp<KspQsp<Ksp, BaSO4BaSO4 will not precipitate in 0.01 mol dm−30.01mol dm−3 H2SO4H2SO4.
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Answered on 10 Apr Learn Equilibrium
Sadika
pH of a solution of a strong acid is 5.0. What will be the pH of the solution obtained after diluting the given solution a 100 times?
Diluting the solution a 100 times means its concentration decreases by a factor of 100. Since it's a strong acid, pH only depends on concentration. Therefore, if the concentration decreases by 100 times, pH increases by 2 units. So, the pH of the diluted solution will be 5+2=75+2=7.
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Answered on 10 Apr Learn Equilibrium
Sadika
The value of KcKc for the reaction 2HI(g)⇌H2(g)+I2(g)2HI(g)⇌H2(g)+I2(g) is 1×10−41×10−4. At a given time, the composition of the reaction mixture is [HI]=2×10−5 mol[HI]=2×10−5mol, [H2]=1×10−5 mol[H2]=1×10−5mol, and [I2]=1×10−5 mol[I2]=1×10−5mol. In which direction will the reaction proceed?
The reaction quotient (QQ) is calculated as [H2][I2][HI]2[HI]2[H2][I2]. If Q<KcQ<Kc, the reaction will proceed in the forward direction to reach equilibrium. If Q>KcQ>Kc, the reaction will proceed in the reverse direction to reach equilibrium. Here, Q=(1×10−5)2(2×10−5)2=14Q=(2×10−5)2(1×10−5)2=41. Since Q<KcQ<Kc, the reaction will proceed in the forward direction.
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Answered on 13/01/2019 Learn Equilibrium
Prabhakar J.
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