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Consider the reaction
H₂(g) + Cl₂(9)→→2HCI(g)
Use the standard thermodynamic data in the tables linked above.
Calculate AG for this reaction at 298.15K if the pressure of HCI(g) is
reduced to 16.30 mm Hg, while the pressures of H₂(g) and Cl₂(g)
remain at 1 atm.
ANSWER: kJ/mol

t t Consider the reaction Hg Cl92HCIg Use the standard thermodynamic data in the tables linked above Calculate AG for this reaction at 29815K if the pressure of class=

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The change in free energy is 19.3kJ.

What is the equilibrium constant?

Now we know that the equation of the reaction is; H₂(g) + Cl₂(g)⇔2HCI(g).

Thus we have that Kp can only be obtained at equilibrium thus we ste up the ICE table as shown;

pH2 = 1 atm

pCl2 = 1 atm

pHCl =  16.30 mm Hg or 0.02 atm

        H₂(g) + Cl₂(g) ⇔ 2HCI(g)

I        1           1                0

C     -x           -x               +2x

E     1 - x         1 -x            2x

When 2x = 0.02

x = 0.01

pH2 =  1 - 0.01

pCl2 = 1 - 0.01

pHCl = 0.02

Kp = (0.02)^2/(0.99)^2

Kp = 4.1 * 10^-4

ΔG = -RT lnKp

ΔG = - (8.314 * 298.15 * ln  4.1 * 10^-4)

= 19.3kJ

Learn more about change in free energy:https://brainly.com/question/14039127

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