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1 year ago

Find ΔG° for the reactions in Problem 20.51 using ΔHf° and S° values.

Chemistry

1 answer:

pashok25 [27]1 year ago

3 0

The value of ΔG° (gibbs free energy change)of the given reaction is -1131.94kj .

Given ,

Balanced chemical equation is given by,

2Mg(s) + O2 (g) → 2MgO (s)

Here , the standard enthalpy of Mg and O2 gas are zero because they are in the most stable state .

thus , the value of ΔG°rxn is given by ,

ΔG°rxn = ΔG°(products ) - ΔG°(reactants )

ΔG°rxn = [2×(-565.97 )] - 0

ΔG°rxn = 2×(-565.97)

ΔG°rxn = -1131.94kj

Hence , the value of ΔG° of the given reaction is -1131.94kj .

<h3>What is a balance chemical reaction ?</h3>

A balanced chemical reaction is a type of reaction which include the reactants and products in same amount i.e. the no of mole on both side of the reaction is remains same or balanced .

Learn more about chemical reaction here :

brainly.com/question/1893305

#SPJ4

Disclaimer :

incomplete question . here is the complete question .

Question :

Calculate ΔG° for the reaction using ΔG°f values :

2Mg(s) + O2 (g) → 2MgO (s)

Find the ΔG° for the reaction using ΔH°f and S° values .

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How many grams of potassium chloride, KCl, must be dissolved in 500.0 mL of solution to produce a 1.5 M solution?

denis23 [38]

Answer:

How many grams of potassium chloride, KCl, must be dissolved in 500.0 mL of solution to produce a 1.5 M solution? Answer: g 4. What is the molarity of a solution in which 84.0 grams of sodium chloride, NaCl, is dissolved in 1.25 liters of solution? Answer: M 5.

Explanation:

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2 years ago

Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. nh4cl cobr3 k2so

Illusion [34]

Answer: CoBr3 < K2SO4 < NH4 Cl

Justification:

1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.

2) The formula for the depression of freezing point is:

ΔTf = i * Kf * m

Where i is the van't Hoof factor which accounts for the dissociation of the solute.

Kf is the freezing molal constant and only depends on the solvent

m is the molality (molal concentration).

3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result in the solution with higher depression of the freezing point (lower freezing point).

4) These are the dissociations of the given solutes:

a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles

b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles

c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles

5) So, the rank of solutions by their freezing points is:

CoBr3 < K2SO4 < NH4 Cl

4 0

3 years ago

Read 2 more answers

Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric

kaheart [24]

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

$pressure\propto \frac{1}{volume}$ (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

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3 years ago

How could you determine if a tissue you are looking at could only be from the heart?

borishaifa [10]

You can determine it by paying attention to the unique characteristics that could only be found at heart's tissue, such as :

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3 years ago

How many liters of O2 gas will be produced at STP from 4 moles of KC1O3

quester [9]

Answer:

V = 134.5 L

Explanation:

Given data:

Number of moles of KClO₃ = 4 mol

Litters of oxygen produced at STP = ?

Solution:

Chemical equation:

2KClO₃ → 2KCl + 3O₂

Now we will compare the moles of KClO₃ with oxygen.

KClO₃ : O₂

2 : 3

4 ; 3/2×4 = 6 mol

Litters of oxygen at STP:

PV = nRT

V = nRT/P

V = 6 mol × 0.0821 atm.L/mol.K × 273 K / 1atm

V = 134.5 L / 1

V = 134.5 L

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3 years ago