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Sunny_sXe [5.5K]
1 year ago
11

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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Answer:

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Explanation:

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3 years ago
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Sodium peroxide (Na2O2) is often used in self-contained breathing devices, such as those used in fire emergencies, because it re
Anastaziya [24]

Answer:

725.15 L

Explanation:

The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:

Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂

From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.

Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:

Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol

moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂

Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol

moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂

Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:

1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂

In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:

1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L

4 0
3 years ago
How to calculate density (what’s the formula) <br> density=_ divided by _
Sphinxa [80]

Answer:

density equals weigt divided by volume

Explanation:

density =  \frac{m}{v}

8 0
3 years ago
What is the volume at stp of 720.0 ml of a gas collected at 20.0 °c and 3.00 atm pressure?
MArishka [77]
Combined gas law is 
       PV/T = K (constant)

P = Pressure
V = Volume
T = Temperature in Kelvin

For two situations, the combined gas law can be applied as,
     P₁V₁ / T₁ = P₂V₂ / T₂

P₁ = 3.00 atm                                     P₂ = standard pressure = 1 atm
V₁ = 720.0 mL                                    T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
  
By substituting,
 3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
                                          V₂ = 2012.6 mL

hence the volume of gas at stp is 2012.6 mL
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