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For election at 500KHX(g)+aq.-----> H⁺ (aq.) + X⁻(aq.)ΔGƒ⁰(X⁻aq) = -131.17 kcal/molΔGƒ⁰(HX,g)= -130.48 kcal/molif at 500K eeuilibrium concentration of [H⁺]=2M &[X⁻]= 2M.what will be equilibrium partial pressure of HX(g) in bar units? [use lnx=2.3 log x]

For election at 500KHX(g)+aq.-----> H⁺ (aq.) + X⁻(aq.)ΔGƒ⁰(X⁻aq) = -131.17 kcal/molΔGƒ⁰(HX,g)= -130.48 kcal/molif at 500K eeuilibrium concentration of [H⁺]=2M &[X⁻]= 2M.what will be equilibrium partial pressure of HX(g) in bar units? [use lnx=2.3 log x]

Grade:12

1 Answers

Vikas TU
14149 Points
5 years ago

In this question he has given the G of formation for X and HX.So we will find the G of the reaction by subtracting 130.48 and 131.17.This gives us G to be -0.69.

 

Now we will use the formula G=-RTlnK to get the value of equilibrium constant.Taking R to be 8.314 ant the temperture as 500K we will get the value of K eo be exactly 1.

 

Since the the phases of the other products are aqueous we simply equate the the partial pressure of HX to the K we got.So the partial pressure is 1.

 

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