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SVETLANKA909090 [29]
3 years ago
8

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

+ 2 O 2 ( g ) What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? Δ G ∘ rxn = kJ/mol What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K? K = What is the equilibrium pressure of O2(g) over M(s) at 298 K? P O 2 = atm
Chemistry
1 answer:
Citrus2011 [14]3 years ago
3 0

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol

b) To calculate the constant we have the following expression:

lnK=-\frac{deltaG_{rxn} }{RT}

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066

c) The equilibrium pressure of O₂ over M is:

K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm

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(Q012) Mafic minerals a. form rhyolite and granite. b. crystallize at a higher temperature than do silicic minerals. c. typicall
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What volume does 2800mg of nitrogen gas occupy at 98Kpa and -10 C?
Evgen [1.6K]

Answer:

2.23L

Explanation:

Using the general gas law as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

Mass of nitrogen = 2800mg

Since 1g = 1000mg

2800mg = 2800/1000

= 2.8g

Using mole = mass ÷ molar mass

Molar mass of nitrogen gas (N2) = 28g/mol

mole = 2.8/28

mole of N2 gas = 0.1mol

Pressure = 98kPa

1 kilopascal (kPa) = 1000pascal (pa)

98kPa = 98 × 1000

= 98000pascal

Since 1 Pascal = 9.869 × 10^-6 atmosphere (atm)

98000pascal = 98000 × 9.869 × 10^-6

= 0.967atm

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Hence, using PV = nRT

0.967 × V = 0.1 × 0.0821 × 263

0.967V = 2.159

V = 2.159/0.967

V = 2.23

Volume of Nitrogen gas = 2.23 L

6 0
3 years ago
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