Calculate the value of Kp for the reaction. Constants| Periodic Table Express yo
ID: 565618 • Letter: C
Question
Calculate the value of Kp for the reaction. Constants| Periodic Table Express your answer to three significant figures A flask is charged with 1.600 atm of N2O4(g) and 1.00 atm of NO2 (g) at 25 C, and the following equilibrium is achieved N3O4 (g) 2NO2(g) After equilibrium is reached, the partial pressure of NO2 is 0.516 atm Submit Previous Answers Request Answer You may want to reference (Pages 639 - 640) Section 15.5 while completing this problem X Incorrect; Try Again; 2 attempts remaining Set up the equilibrium expression for this reaction and use the given equilibrium pressure o the product (0.516 atm) and the equilibrium pressure of the reactant (determined in Part A) to calculate Kp Part C Calculate the value of Kc for the reaction. Express your answer to three significant figures.Explanation / Answer
Data given.
PN2O4= 1.6 atm
PNO2 = 1atm
In equilibrium:
PN2O4= 1.6 -x
PNO2 = 1 + 2x
and PNO2 = 1 + 2x = 0.516
x = (0.516-1)/2 = -0.242
Then
PN2O4= 1.6 -x = 1.6 - (-0.242) = 1.842 atm
b)
Kp = PNO2^2 / PN2O4 = (0.516) ^2/1.842 = 0.1445
c)
Kp = Kc*(RT)^dn
Kc = Kp / (RT)^dn
Kc = (0.1445)/((.082*298)^(2-1)) = 0.005915
Kc = 0.006
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.