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=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:

1. At T1=298 KT1=298K: K1=e−118.3×1038.314×298K1=e−8.314×298118.3×103

2. 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.

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:

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.