Question

* I unerstand this is a chemical engineering question but that was not a subject choice *

Using the information in Perry's Chemical Engineer's Handbook, compute the Henry's law constant in atm for nitrogen in water at 299.0 degrees Kelvin.

I have access to the values in Perry's Chemical Engineering Handbook, I just have absolutely no idea how to solve this.

Answer 1

The Henry´s law says that

c= K_H P

where c is the concentration of the gas,  K_H is the constant of Henry´s law and P is the the parcial presure.

Now we want to calculate the constant of Henry´s law

K_H =c/P

Now the molarity is the mole of sulute/Volumen of solution and P is the parcial pressure is Pi= XiP_t where Xi is the molar fraction of the solute and P_t is the total pressure. Now we subsitute in the equation

K_H = (mole sto/Vsolution)/XiP_t

Now the molar fraction is Xi=mole sto/ total mole in solution, now we substitute again:

K_H = (mole sto/Vsolution)/[(mole sto/total mole) P_t ]

We see that we have mole sto in the numerator and denominator so we can eliminate it.

K_H = (1/Vsolution)/[(1/total mole) P_t ]

Now with the ideal gas equation we can substitute the value of pressure P_t =nRT/V

Where n is the total mole and the Volumen is the total volume so is the volume of the solution, and we substitute again:

K_H = (1/Vsolution)/[(1/total mole)nRT/V ]

So the Vsolution in the numerator can be eliminate it with the V in the denominator, and total mole can be eliminate it with n, so the equation final is:

K_H = (1/1)/[(1/1)RT/1 ] appliying the doble c you get:

K_H = RT

so the henry´s law is the constant of the ideal gases by the temperature in Kelvin. Because they are asking the constant in atm we are going to use the value of 0,082Latm/Kmole

K_H = 0,082Latm/Kmol *299,0K = 24,52 Latm/mol