2- For the reaction 2NO(g) + O2(g)2NO2(g) H° = -114.2 kJ and S° = -146.5 J/K...

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2- For the reaction 2NO(g) + O2(g)2NO2(g) H° = -114.2 kJ and S° = -146.5 J/K...

2- For the reaction 2NO(g) + O2(g)2NO2(g) H° = -114.2 kJ and S° = -146.5 J/K The equilibrium constant for this reaction at 289.0 K is ---------- . Assume that H° and S° are independent of temperature.

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Answer 1

Answer #1

ΔHo = -114.2 KJ/mol

ΔSo = -146.5 J/mol.K

= -0.1465 KJ/mol.K

T = 289 K

use:

ΔGo = ΔHo - T*ΔSo

ΔGo = -114.2 - 289.0 * -0.1465

ΔGo = -71.8615 KJ/mol

T = 289 K

ΔGo = -71.8615 KJ/mol

ΔGo = -71861.5 J/mol

use:

ΔGo = -R*T*ln Kc

-71861.5 = - 8.314*289.0* ln(Kc)

ln Kc = 29.9081

Kc = 9.748*10^12

Answer: 9.75*10^12

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Answer 2

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