The unbalanced chemical equation with the corresponding phase for each substance is shown below:

Cu₍s₎ + H₂SO₄₍aq₎ → CuSO₄₍aq₎ + SO₂₍g₎ + H₂O₍l₎

The copper metal is solid, the sulfuric acid and copper sulfate are in aqueous solution. Water is in the liquid phase while sulfur dioxide is in the gas phase.

In order to obtain the balanced equation, we must ensure that the number of atoms present in the reactant side is equal to that in the product side. First, we list the elements involved and count their number on each side.

Reactant Product
1 Cu 1
2 H 2
1 S 2
4 O 5

To balance the equation, we need to have 2 moles of H₂SO₄ and 2 moles of H₂O. The balanced chemical equation is then:

Cu₍s₎ + 2H₂SO₄₍aq₎ → CuSO₄₍aq₎ + SO₂₍g₎ + 2H₂O₍l₎

Reactant Product
1 Cu 1
4 H 4
2 S 2
8 O 8

3 0

3 years ago

The normal boiling point of ethanol (c2h5oh) is 78.3 °c and its molar enthalpy of vaporization is 38.56 kj/mol. what is the chan

kaheart [24]

When the entropy change of vaporization

= enthalpy of vaporization/boiling point temperature

when we have the enthalpy of vaporization = 38560 J/mol

and the boiling point temperature in Kelvin = 78.3 + 273 = 351.3 K

by substitution:

∴the entropy change of vaporization = 38560J/mol/351.3K

= 109.76 J/K/mol

and when the liquid has lesser entropy than the gas and we here convert from

gas to liquid so, the change in entropy = -109.76 J/K/mol

now, we need the moles of C2H5OH = mass/molar mass when the molar

mass of C2H5OH = 46 g/mol and mass = 42.2 g

∴ moles of C2H5OH = 42.2 g / 46 g/mol = 0.92 moles

when 1 mol of C2H5OH turns in liquid entropy change →-109.76 J/K/mol

∴ 0.92 mol of C2H5OH → X

∴ X entropy change when 0.92 mol = -109.76 *0.92 mol / 1 mol

= 84.64 J/K

6 0

3 years ago

Consider the Gibbs energies at 25 ∘C.

zysi [14]

Answer:

A)ΔGrxn∘=55.7kJ/mol

B)Ksp=1.75×10^−10

C)ΔGrxn∘=70.0kJ/mol

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=ΔGf,products∘−ΔGf,reactants∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Cl(aq)−∘]−ΔGf,

AgCl(s)⇌Ag(aq)++Cl(aq)−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

ΔGrxn∘=55.7kJ/mol

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7kJ/mol=−(8.3145×10−3J/molK)(298.15K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Br(aq)−∘]−ΔGf,

ΔGrxn∘=[77.1kJ/mol+(−104.0kJ/mol)]−(−96.90kj/mol

ΔGrxn∘=70.0kJ/mol

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K

Ksp=5.45×10^−13

8 0

3 years ago