A flask containing 0.158 M methanol CH_3OH, and 0.172 M oxygen gas is allowed to

Question

A flask containing 0.158 M methanol CH_3OH, and 0.172 M oxygen gas is allowed to come to equilibrium at 300 k. At equilibrium the flask is found to contain 0.098 M oxygen gas determine the equilibrium constant (kc) for this reaction. 2CH_3OH(g) + 3O_2(g) doubleheadarrow 2CO_2(g) = 4H_2O(g) 0.0207 1.79 Times 10^3 1.30 0.0138 25.3 The equilibrium constant, K_c = 0.0563 for the reaction shown below. The enthalpy change for this reaction at 25 degree C is -25.6 kJ. 2 A + B rightarrow 3C What is the correct value of Kc for the reaction shown below? 3/2 C rightarrow A + 1/2 B 8.88 0.0282 4.21 0.237 17.8 CH_3COCH_3 (acetone) is a commonly used organic solvent. Determine the standard entropy change (Delta S degree rxn) in J/K for the following reaction of acetone: CH_3COCH_3 (l) + 4O_2 (g) rightarrow 3CO_2 (g) + 3H_2O (g) If we know S degree values for CH_3COCH_3 (l), O_2(g), CO_2 (g) and H_2O (g) are 198.7 J/Kmol, 205.0 J/Kmol, 213.6 J/Kmol,, and 188.7 J/Kmol,respectively. -1.40 2225. 188.2 806.0 -188.6

Explanation / Answer

19)

consider the given reaction

2 CH3OH + 3 02 ---> 2 C02 + 4 H20

initial concentrations are

[CH3OH] = 0.158

[02] = 0.172

now

using ICE table

at equilibrium

[CH3OH] = 0.158 - 2x

[02] = 0.172 - 3x

[C02] = 2x

[H20] = 4x

given

at equilibrium , [02] = 0.098 M

so

0.172 - 3x = 0.098

x = 0.024666

so

[CH3OH] = 0.158 - ( 2*0.024666) = 0.1086666

[02] = 0.098

[C02] = 2 * 0.024666 = 0.049333

[H20] = 4 * 0.024666 = 0.0986666

now

Kc = [C02]^2 [H20]^4 / [CH3OH]^2 [02]^3

Kc = [0.049333]^2 [0.098666]^4 / [0.108666]^2 [0.098]^3

Kc = 0.020702

so

the answer is option A) 0.0207

20)

consider the first reaction

2 A + B ---> 3C

the equilibrium constant is given by

Kc1 = [C]^3 / [A]^2 [B]

now

consider the second reaction

1.5 C --> A + 0.5 B

the equilibrium constan tis given by

Kc2 = [A] [B]^0.5 / [C]^1.5

from the equations of Kc1 and Kc2

we can see that

Kc1 = 1 / ( Kc2)^2

so

0.0563= 1/ (Kc2)^2

Kc2 = 4.21

so

the answer is option C) 4.21

8)

we know that

dSo rxn = So products - So reactants

so

dSo rxn = ( 3 x So C02) + ( 3 x So H20) - So CH3COCH3 - ( 4 x So 02)

using givne values

we get

dSo rxn = ( 3 x 213.6) + ( 3 x 188.7) - 198.7 - ( 4 x 205)

dSo rxn = 188.2

so

entropy change for the reaction is 188.2

so

the answer is option C) 188.2

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