Question

Determine the value of Kp for the following reaction if the equilibrium concentrations are as follows: P(CO)eq = 6.8 × 10-11 atm, P(O2)eq = 1.3 × 10-3 atm, P(CO2)eq = 0.041 atm.

2 CO(g) + O2(g) ⇌ 2 CO2(g)

A) 3.6 × 10-21
B) 2.8 × 1020
C) 4.6 × 1011
D) 2.2 × 10-12
E) 3.6 × 10-15
Answer: B

Answer 1

I know how to find Kc but i cant figure out how to solve for Kp

Answer 2

use product over reactant, no solids or liquids

Answer 3

Il work it out 1sec, its been a while since chem 2

Answer 4

kp= P^2 [CO2] / P^2[ CO2] P [O2]

Answer 5

ok

Answer 6

Perfect yes, just plug and solve

Answer 7

Thats it

Answer 8

yep, wait till you take analytic chem there is more steps

Answer 9

With the P^2 would that mean square it

Answer 10

yes

Answer 11

ok thanks very much

Answer 12

np anymore questions feel free to ask

Answer 13

yes

Answer 14

when do we know the kp=Kc

Answer 15

1 sec let me use the rr

Answer 16

ok

Answer 17

For the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), Kc = 8.3 × 10-10 at 25°C. What is the concentration of N2 gas at equilibrium when the concentration of NO2 is five times the concentration of O2 gas?
A) 3.3 × 10-11 M
B) 1.7 × 10-10 M
C) 6.0 × 109 M
D) 3.0 × 1010 M
Answer: D

Answer 18

i suppose i have to start another question

Answer 19

kk, kc is used for equilibrium and molaritys, kp is used for solubility ksp

Answer 20

so when we are using delta n,
when both sides are equal thats when kp=kc

Answer 21

yes

Answer 22

kp is pressure and precipitation, kc-moles

Answer 23

ok

Answer 24

can you pull your second problem up without the symbols

Answer 25

it did that by itself so i will close this one and start another question

Answer 26

kk