Question

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 1

for this one would we use

kc= kp/ (RT)^delta n

Answer 2

since we are looking for the gas it would be
k[=Kc(RT) delta n

Answer 3

Kp for the first

Answer 4

kk, you would use that if given kc to find ksp,

Answer 5

let me work it out real quickly, i do not want to tell you wrong

Answer 6

ok

Answer 7

kk

Answer 8

1.) kc=(NO)^2/(N2)(O2)2

2.) 8.3*10^-10=(5)^2/(N2)(x)2

3.) 8.3*10^-10=25/N2
N2=3.0*1010M

Answer 9

Does it make sense?

Answer 10

i get step one
for part 2 i see why you plugged in kc
but i dont understand everything else

Answer 11

kk, We are solving it like an algebraic expression
we know kc value, and it is asking what is N2(M)\
we know equilibrium is products/ reactants
kc=product/reactants
we were given value for NO2=5, so O2 was just 1, because it stated NO2 was 5 times of O2

so kc=8.3*10^-10, NO2=5, O2=1, N2= we are solving for
8.3*10^-10=(NO2=5)^2/(N2)(O2=1)^2
so rearrange the equation to say
N2=25/8.3*10^-10= 3.0*10^-10

Answer 12

ok that makes a lot more sense, at first i didnt understand why o2 just left

Answer 13

Aweseome, for chem 2 you can always make the other variable 1 in equilibrium

Answer 14

do you have time for one more

Answer 15

Yeah, Im on fall break

Answer 16

yayyyyy lets do this okay i will start another question

Answer 17

kk