Question

Can I get chemistry help?

Answer 1

@AZ

Answer 2

I have no idea for #8
I want to say there's something wrong with the answer choices given

For #9, they gave us the equation
\(\sf 2~SO_2(g) + O_2(g) \rightleftharpoons 2 ~SO_3(g)\)

So if you have the reaction
\( aA + bB \rightleftharpoons cC + dD\)

Then \(\sf K_p\) can be calculated using the partial pressures of the reactants and products

\( K_p = \dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \)


And so for your question

\(\sf K_p = \dfrac{(P_{SO_3})^2}{(P_{SO_2})^2(P_{O_2})}\)

Plug in the values they gave you
\(\sf K_p = 0.345\)
\(\sf P_{SO_2} = 35\)
\(\sf P_{O_2} = 15.9\)

and solve for \(\sf P_{SO_3}\)

Answer 3

82 atm

Answer 4

You got it

Answer 5

Okay if you could help me with one more and Ill be done for the day haha

Answer 6

Which of the following statements is true?
Group of answer choices

An equilibrium can be established only from reactants

An equilibrium can be established from any combination of products and rectants

An equilibrium can be established only from products

An equilibrium can be established only from an equimolar mixture of products and reactants

Answer 7

So what do you think it is?

What do you know about reactions reaching equilibrium?
Do you need to have only reactants or only products? Can it reach equilibrium with both?

Answer 8

I think its the last choice

Answer 9

equimolar means you're adding the same amount of moles

think about it

you have a reaction
x \(\leftrightharpoons\) y

you add a bunch of x, and the reaction will move to the right to get to equilibrium
you add a bunch of y, and the reaction will move to the left to get to equilibrium

Answer 10

So the second choice?

Answer 11

D is way too specific and not true because

yup, you got it :)

Answer 12

Thank you

Answer 13

No problem!

Answer 14

Now on the question that we couldn't figure out. I thought there would be an answer choice with more moles going to N2O4 than NO2 which would make it go right but there isn.t

Answer 15

The only thing that may have a piece for that is the equilibrium constant of 4 of the professor just had a typo.

Answer 16

@AZ

Answer 17

Or it could be the last answer choice just because 0.50 x 4 is 2.00 but I'm not sure why the constant means

Answer 18

OHHHHHH thank you, your last comment made the lightbulb go off in my head

Answer 19

we have to calculate \(\sf Q_{c}\)

and then compare that to \(\sf K_{eq}\) which is equal to 4

\( aA +bB \rightleftharpoons cC +dD\)

\(Q_{c}= \dfrac{[C]_{init}^c[D]_{init}^d}{[A]_{init}^a\text{[}B]_{init}^b} \)

Answer 20

so for each answer choice they gave you the moles, and it's in a 1 L vessel

basically moles / liter gives you the concentration

and since it's 1 L that means those numbers are the concentration

Answer 21

So it would be the first option?

Answer 22

So calculate the \(\sf Q_c\) for each answer choice

if
\(\sf Q_c > K\) then it shifts to the left

if
\(\sf Q_c < K\) then it shifts to the right

Answer 23

\(\color{#0cbb34}{\text{Originally Posted by}}\) @kamachavis
So it would be the first option?
\(\color{#0cbb34}{\text{End of Quote}}\)
no

for your reaction, you have

\( \sf Q_c = \dfrac{[NO_2]^2}{[N_2O_4]}\)

plug in those values and then compare \(\sf Q_c\) with K which is 4

Answer 24

the NO2 value gets squared

Answer 25

So third option

Answer 26

That's your answer!