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katrin [286]
3 years ago
8

Find the equilibrium constants, Kp, for the following equilibria, (i) NO(g) + ½ O2(g) ⇄ NO2(g), Kp = ? (ii) NO2(g) ⇄ NO(g) + ½ O

2(g), Kp = ?, given the equilibrium constant, Kp, for the reaction: 2NO (g) + O2(g) ⇄ 2NO2(g) Kp= 100 at the same temperature
Chemistry
1 answer:
Roman55 [17]3 years ago
3 0

Answer:

Equilibrium constant for the 1st reaction is 10 and for the 2nd reaction is 0.1

Explanation:

2NO+O_{2}\rightleftharpoons 2NO_{2}

K_{p}=\frac{P_{NO_{2}}^{2}}{P_{NO}^{2}\times P_{O_{2}}}=100

(i) NO+\frac{1}{2}O_{2}\rightleftharpoons NO_{2}

Equilibrium constant, K_{p1}=\frac{P_{NO_{2}}}{P_{NO}\times P_{O_{2}}^{\frac{1}{2}}}=\sqrt{K_{p}}=\sqrt{100}=10

(ii) NO_{2}\rightleftharpoons NO+\frac{1}{2}O_{2}

K_{p2}=\frac{P_{NO}\times P_{O_{2}}^{\frac{1}{2}}}{P_{NO_{2}}}=\frac{1}{K_{p1}}=\frac{1}{10}=0.1

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Leni [432]

Answer:

a) \Delta G=2.6kJ

b) \Delta G=-979.57kJ

c) \Delta G=264.21kJ

Explanation:

Hello,

In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:

\Delta _fG_{H_2}=\Delta _fG_{I_2}=0kJ/mol\\\Delta _fG_{HI}=1.3kJ/mol\\\Delta _fG_{CO_2}=-394.4kJ/mol\\\Delta _fG_{CO}=-137.3 kJ/mol\\\Delta _fG_{NH_3}=16.7 kJ/mol\\\Delta _fG_{HCl}=-95.3kJ/mol\\\Delta _fG_{MnO_2}=465.37kJ/mol\\\Delta _fG_{Mn}=0kJ/mol\\\Delta _fG_{NH_4Cl}=-342.81kJ/mol

So we proceed as follows:

a)

\Delta G=2\Delta _fG_{HI}-\Delta _fG_{H_2}-\Delta _fG_{I_2}\\\\\Delta G=2*1.3\\\\\Delta G=2.6kJ

b)

\Delta G=\Delta _fG_{Mn}+2*\Delta _fG_{CO_2}-\Delta _fG_{MnO_2}-2*\Delta _fG_{CO}\\\\\Delta G=0+2*-394.4-465.37-2*-137.3\\\\\Delta G=-979.57kJ

c)

\Delta G=\Delta _fG_{NH_3}+\Delta _fG_{HCl}-\Delta _fG_{NH_4Cl}\\\\\Delta G=16.7-95.3-(-342.81)\\\\\Delta G=264.21kJ

Regards.

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