Ask question

Ask question

All categories

Natasha2012 [34]

2 years ago

8

A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.9

56 g lead (II) bromide (PbBr2 solid at a temperature of 25.0°C. Under these conditions, calculate the reaction free energy AG for the following chemical reaction: Pb2+ (aq) + 2Br (aq) = PbBr2 (s) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

1 answer:

Ronch [10]2 years ago

8 0

Answer:

The free energy = -20.46 KJ

Explanation:

given Data:

Pb²⁺ = 0.750 M

Br⁻ = 0.232 M

R = 8.314 Jk⁻¹mol⁻¹

T = 298K

The Gibb's free energy is calculated using the formula;

ΔG = ΔG° + RTlnQ -------------------------1

Where;

ΔG° = standard Gibb's freeenergy

R = Gas constant

Q = reaction quotient

T = temperature

The chemical reaction is given as;

Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)

The ΔG°f are given as:

ΔG°f (PbBr₂) = -260.75 kj.mol⁻¹

ΔG°f (Pb²⁺) = -24.4 kj.mol⁻¹

ΔG°f (2Br⁻) = -103.97 kj.mol⁻¹

Calculating the standard gibb's free energy using the formula;

ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)

Substituting, we have;

ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]

ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)

= -260.75 + 232.34

= -28.41 kj

Calculating the reaction quotient Q using the formula;

Q = 1/[Pb²⁺ *(Br⁻)²]

= 1/(0.750 * 0.232²)

= 24.77

Substituting all the calculated values into equation 1, we have

ΔG = ΔG° + RTlnQ

ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)

= -28.41 +7.95

= -20. 46 kJ

Therefore, the free energy of reaction = -20.46 kJ

You might be interested in

Consider the following balanced redox reaction: 2CrO2-(aq) + 2H2O(l) + 6ClO-(aq) LaTeX: \longrightarrow⟶ 2CrO42-(aq) + 3Cl2(g) +

frutty [35]

Answer:

1. Chromium

2. Chlorine.

3. Chlorine.

4. Chromium.

5. 12 electrons.

Explanation:

Hello,

In this case, the given reaction with the appropriate oxidation states turns out:

$2(Cr^{+3}O^{-2}_2)^-(aq) + 2H_2O(l) + 6(Cl^{+1}O^{-2})^-(aq)\longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq) + 3Cl^0_2(g) + 4OH^-(aq)$

In such a way, the oxidation half-reaction is written for chromium as the reducing agent so it is oxidized from +3 to +6, nonetheless, since there are two chromiums undergoing such change, 6 electrons are being transferred as shown below:

$2(Cr^{+3}O^{-2}_2)^-(aq) \longrightarrow 2(Cr^{+6}O^{-2}_4)^{2-}(aq)+6e^-$

On the other hand, chlorine's reduction half-reaction as the oxidizing agent result from the transfer of 6 electrons as well from +1 to 0, nonetheless, there are 6 chlorines undergoing such change:

$6(Cl^{+1}O^{-2})^-+6e^-\longrightarrow 3Cl^0_2(g)$

Therefore, there are 12 electrons that are being transferred, 6 for chromium and 6 for chlorine.

Best regards.

5 0

2 years ago

The weight percent of concentrated HClO4(aq) is 70.5% and its density is 1.67 g/mL. What is the molarity of concentrated HClO4

ollegr [7]

Answer:

[HClO₄] = 11.7M

Explanation:

First of all we need to know, that a weight percent represents, the mass of solute in 100 g of solution.

Let's convert the mass to moles → 70.5 g . 1mol/100.45 g = 0.702 moles

Now we can apply the density to calculate the volume.

Density always refers to solution → Solution density = Solution mass / Solution volume

1.67 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.67 g/mL → 59.8 mL

To determine molarity (mol/L) we must convert the mL to L

59.8 mL . 1L/1000mL = 0.0598 L

Molarity → Moles of solute in 1L of solution → 0.702 mol / 0.0598 L = 11.7M

8 0

2 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

B)

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

2 years ago

In science class, Blaine’s teacher puts one glow stick in a cup of hot water and another glow stick in a cup of cold water. She

timofeeve [1]

Answer:

The glow stick in hot water will be brighter

Explanation:

The glow stick in hot water will be brighter than the glow stick in cold water because the heat from the hot water will cause the molecules in the glow stick to move faster. The faster the molecules move in the glow stick, the sooner and brighter the reaction will be. The cold water will cause molecules to move slowly and it will take longer for the reaction to occur, which will also make it less bright.

3 0

2 years ago

Read 2 more answers

If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate

andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

6 0

2 years ago