1 mol N2O4 has ?P. Use PV = nRT. I obtained approximately 5 atm but you need to do it exactly.
2 mol NO2 = ?P. Same thing. I get approximately 10 atm.
............N2O4 ==> 2NO2
initial.....5..........10
change......+p..........-2p
equl.......5-p..........10-2p
How do you know which way the reaction will go. That is done by the equilibrium quotient.
Kquo = pNO2^2/pN2O4 = 10^2/5 = about 20. Compare that with 0.1134 and Kquo is too large which means the numerator is too high and the denominator is too low. So the reaction must go to the left to reach equilibrium.
Kp = 0.1134 = pNO2^2/pN2O4
Substitute from the ICE chart above and solve for p. Add that to the initial p to find final p. Then substitute that p into PV = nRT and solve for n. That will give you moles N2O4 at equilibrium and n/5L will give you molarity. I worked the problem and I obtained one of the possible answers.
Consider the equilibrium system: N204 (g) = 2 NO2 (g) for which the Kp = 0.1134 at 25 C and deltaH rx is 58.03 kJ/mol. Assume that 1 mole of N2O4 and 2 moles of NO2 are introduced into a 5 L contains. What will be the equilibrium value of [N204]?
Options are:
A) 0.358 M
B) 0.042 M
C) 0.0822 M
D) 0.928 M
E) 0.379 M
Please provide an explanation, I want to understand this question.. thank you so much!!
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