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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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2 years ago
A transformer has 90 turns in the primary coil and 9 turns in the secondary coil If the output voltage is 6 V, what is the input
Afina-wow [57]

Answer:

60 V

Explanation:

From;

Vs/Vp = Ns/Np

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6 0
3 years ago
Calculate density of aluminum if 27.6cm^3 has a mass of 74.6g
rjkz [21]
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6 0
3 years ago
Read 2 more answers
Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe
Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

V_s=V_o-V_i=V_i\\V_o=2*V_i\\

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

r_o=r_i+e

The first equation becomes

\frac{4 \pi}{3}*r_o^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\\frac{4 \pi}{3}*(r_i+e)^{3}  = 2 * \frac{4 \pi}{3}*r_i^{3}  \\\\(r_i+e)^{3}  = 2 *r_i^{3}  \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

8 0
2 years ago
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