Chem 2: Molecular Equilibria

[CausativeAgent]

Homework Statement

The standard Gibb's free energy for the reaction N2+3H2<--->2NH3 is -33.0 KJ at 298 K. Everything is a gas. Calculate Kp for this reaction.

Homework Equations

I have in my notes that Keq=e^(-Delta G/RT) where R is the constant 0.008314 KJ/K. I also have a formula that converts between Kp and Keq: Kp= Kc(.0821Xtemperature)^(Moles of gasseous products-moles of gasseous reactants).

The Attempt at a Solution

My attempt at a solution was to use the first formula (which I think gives Kc) and then convert it into Kp using the second formula.
My answer ended up differing from the one given in my homework though. Here is my work:

Keq= e^(33.0/(0.008314x298)= 6.09x10^5

Kp= (6.09x10^5)(0.0821x298)^(-2) = not the correct answer

The answer is: Kp=5.97x10^5

 

[chaoseverlasting]

I think here, you have to use R=8.314 joules/mol . Keq can be calculated in the same way as you showed.

That is Kc. Kp=Kc(RT)^(delta Ng)

 

[Blueshift5]

Your approach was correct, but there may have been a calculation error in converting from Keq to Kp. Here is the correct solution:

1. Calculate Keq using the first formula:
Keq = e^(-33.0/(0.008314*298)) = 6.09x10^-6 (note the negative exponent)

2. Use the second formula to convert from Keq to Kp:
Kp = (6.09x10^-6)(0.0821*298)^(2-4) = 5.97x10^5

Your answer differed because you used a positive exponent for the temperature in the second formula, which should be negative. Also, the moles of gaseous products and reactants were not specified in the problem, so you should leave them as variables in the formula. Hope this helps!