-4.73kJ/mol at 298 K for the reaction 3. ArG 2NO2(g)N2O4(g) If the initial pressure of pNO,...

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science chemistry

Answer 1

Answer #1

$\Delta$ G⁰ = -4.73KJ/mole

= -4730J/mole

$\Delta$ G⁰ = -RTlnKp

-4730 = -8.314*298lnJp

lnKp = -4730/(-8.314*298)

lnKp = 1.91

Kp = 6.753

-------- 2NO2(g) --------------> N2O4(g)

I ---- 2*10^4 ---------------- 0

C ----- -2x ------------------ x

E ---- 2*10^4-2x ----------- x

Kp = PN2O4/P^2NO2

6.753 = x/(2*10^4-2x)

6.753*(2*10^4-2x) = x

x eq = 9310Pa

PN2O4 = 9.31*10^3 Pa >>>>answer

Answer 2

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