How many moles/liter of NOBr must be added to the above equilibrium mixture to produce an equilibrium mixture that is 0.450 M Br2?
How do I do b) from this set of questions?
Consider the following equilibrium:2NOBr(g) --> 2NO(g) + Br2(g)An equilibrium mixture is 0.160 M NOBr, 0.239 M NO, and 0.205 M Br2.a) What is the value of Kc at the temperature of the above concentrations?Kc = 0.457Mb) How many moles/liter of NOBr must be added to the above equilibrium mixture to produce an equilibrium mixture that is 0.450 M Br2?__mol/L NOBr must be addedc) If the temperature is 390 K, what is the value of Kp?Kp = 14.6 atmd) What is the value of Go at 390 K?dGo = -8.7 kJ
- hcbiochemLv 71 month ago
Let me try again with a different approach.
a) Kc = [NO]^2[Br2]/[NOBr]^2
Kc = (0.239)^2(0.205) / (0.160)^2 = 0.457
b) When [Br2] increases to 0.450, it has increased by 0.245 M. Because of the stoichiometry of the reaction, [NO] will increase by twice as much (by 0.490) to 0.729 M.
Now, at this new equilibrium, calculate [NOBr]:
Kc = 0.457 = (0.245)(0.729)^2 / [NOBr]^2
[NOBr] = 0.284 M
But we have to have added sufficient NOBr to form 0.490 M of NO, so before any NOBr decomposed, [NOBr] = 0.490 + 0.284 = 0.774 M
The other two parts were correct:
c) The relationship between Kc and Kp is given as:
Kp = Kc (RT)^∆n
Where ∆n = moles of gaseous products - moles of gaseous reactants
Kp = 0.457 (0.08206 X 390 K)^(3-2)
Kp = 14.6
d) ∆G° = - RT ln Kc
∆G° = -8.314 J/molK (390 K) ln (14.6) = -8700 J/mol = -8.70 kJ/mol