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Maksim231197 [3]
3 years ago
12

Consider a reaction that has a positive ΔH and a positive ΔS. Which of the following statements is TRUE?A) This reaction will be

spontaneous only at high temperatures.B) This reaction will be spontaneous at all temperatures.C) This reaction will be nonspontaneous at all temperatures.D) This reaction will be nonspontaneous only at high temperatures.E) It is not possible to determine without more information.
Chemistry
1 answer:
Alja [10]3 years ago
4 0

Answer: A) This reaction will be spontaneous only at high temperatures

Explanation:

\Delta G= +ve, reaction is non spontaneous

\Delta G= -ve, reaction is spontaneous

\Delta G= 0, reaction is in equilibrium

Using Gibbs Helmholtz equation:

\Delta G=\Delta H-T\Delta S

Given : \Delta H=+ve

\Delta S=+ve

\Delta G=(+ve)-T(+ve)

\Delta G=(+ve)(-ve)

Thus the value of  \Delta G  is negative and spontaneous when temperature is high.

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Answer:

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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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