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2 years ago

In a closed system, how will a decrease in pressure affect the following reaction: N2O4(g) ⇌ 2NO2(g)?

Chemistry

1 answer:

timama [110]2 years ago

5 0

Answer:

Answer 'A'

Explanation:

In general, if the Σmolar volumes(g) reactants ≠ Σmolar volumes(g) products, a change in atmospheric pressure will shift the reaction equilibrium. If the pressure is increased, the rxn will shift toward the LOWER molar volume side of the rxn or if the pressure is decreased the rxn will shift toward the HIGHER molar volume side. For the reaction N₂O₄(g) ⇄ 2NO₂(g), Vm(N₂O₄(g)) < Vm(NO₂(g)) so, a decrease in atmospheric pressure would shift rxn toward the NO₂(g) side of the equation increasing the moles of NO₂(g). Also, note that if the Σmolar volumes(g) reactants = Σmolar volumes(g) products, no shift in equilibrium will occur regardless of changes in atmospheric pressure.

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3 years ago

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3 years ago

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mafiozo [28]

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2 years ago

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We're only talking about molar mass and time (t) here so we'll just concentrate on $\sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$ . Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

$\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}$

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

$(\sqrt{\frac{M2}{131} } )^2= (\frac{2.29}{4.83})^2$

${\frac{M2}{131} } = (0.47)^2$

${\frac{M2}{131} } = 0.22$

M2= 0.22 x 131

M2= 28.93 g/mol

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2 years ago