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Evgen [1.6K]
3 years ago
9

Given that Kp [NOTE: Kp!!!!] = 1.39 at 400 ºC for the reaction, P4(g) <=> 2 P2(g), which answer best describes the reactio

n when 2.50 mol of P4(g) and 1.50 mol of P2(g) are mixed in a 25.0 L, closed container at 400 ºC? [REMEMBER: it’s Kp]
Chemistry
1 answer:
Igoryamba3 years ago
3 0

Answer:

The reaction will proceed to the left to attain equilibrium.

Explanation:

The question is missing but I guess it must be about <em>how the reaction will proceed to attain equilibrium.</em>

First, we have to calculate the partial pressures using the ideal gas equation.

pP_{4}=\frac{2.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L} =5.52atm

pP_{2}=\frac{1.50mol\times (0.08206atm.L/mol.K)\times 673K}{25.0L}=3.31atm

Now, we have to calculate the reaction quotient (Qp).

Qp=\frac{pP_{2}^{2}}{pP_{4}} =\frac{3.31^{2} }{5.52} =1.98

Since Qp > Kp, the reaction will proceed to the left to attain equilibrium.

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3 years ago
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HELP Which of the following fractions can be used in the conversion of 32 m3 to the unit mm3?
poizon [28]
We know that each millimeter contains 10⁻³ meters. Writing this as a ratio:
1 mm : 10⁻³ m

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Which of the data sets qualitative data?
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3 0
3 years ago
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A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

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3 years ago
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