• Consider the reaction N2O4 (g) ⇌ 2 NO2 (g). At equilibrium, a 2.00-L reaction vessel contains NO2 at a pressure of 0.269 atm and N2O4 at a pressure of 0.500 atm. The reaction vessel is then compressed to 1.00 L. What will be the pressures of NO2 and N2O4 once equilibrium is re-established?

Explanation:

Hello,

At first, it is convenient to compute Kp, since it remains constant during the proposed situation:

Now, since the volume is halved, the pressures of N₂O₄ and NO₂ are doubled, 1.00 atm and 0.538 atm respectively, but now they are considered as the initial pressures, so a change is now present in order to take the reaction again to the equilibrium. Besides, the reaction changes as the products have more moles as:

Thus, the law of mass action goes (considering the reverse reaction):

Hence, solving for by using solver or the quadratic equation, one obtains:

Nonetheless, the feasible solution is 0.072 atm as 0.50 atm leads to a negative pressure of NO₂, thus, the new equilibrium pressures are:

Best regards.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g