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77julia77 [94]
3 years ago
11

The reaction A( g ) ⇌ 2 B( g ) A(g) ⇌ 2 B(g) has an equilibrium constant of K = 0.010 K = 0.010. What is the equilibrium constan

t for the reaction B( g ) ⇌ 1 2 A( g ) ?
Chemistry
1 answer:
Over [174]3 years ago
5 0

Answer:

K = 10

Explanation:

Using Hess's law, it is possible to obtain the equilibrium constant, K, of a reaction using K of similar reactions. For example:

<em> If A ⇄ B K = X</em>

B ⇄ A K = 1/X

2A ⇄ 2B K = X².

Thus, if A(g) ⇄ 2B(g) K = 0.010

2B(g) ⇄ A(g) K = 1 / 0.010; K = 100

B(g) ⇄ A(g) K = √100 = 10

<h3>K = 10</h3>
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A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
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This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

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