Question

For the following reaction, the partial pressures are listed in the table. If the Kp of the reaction is 0.134, which direction would the reaction need to go to establish equilibrium?
H2S(g) + I2(s) <--> 2HI(g) + S(s)
https://learning.k12.com/d2l/common/viewFile.d2lfile/Database/NDM3MzIxMw/27f04366-076d-4f5c-843d-247e476de854.jpg?ou=110233

Answer 1

Since Q < Kp, the reaction must proceed to the right.


Q = Kp so the reaction is at equilibrium.


Since Q < Kp, the reaction must proceed to the left.


Since Q > Kp, the reaction must proceed to the left.

Answer 2

@Photon336 @Prathamesh_M

Answer 3

I'm thinking C.

Answer 4

What do you guys think? @Photon336 @Prathamesh_M @sweetburger

Answer 5

Can you help me with this one? @Photon336 @Prathamesh_M

Answer 6

Kp would include only partial pressures right?

Answer 7

since two components of the given rxn are solids i don't think u should take them into account

Answer 8

consider only the gases when u calculate Q

Answer 9

Okay well how do I calculate Q

Answer 10

@Prathamesh_M

Answer 11

\[\frac{ [HI]^{2} }{ [H _{2}S] }\]

Answer 12

Okay so I got an answer, then where do I go from there? @Prathamesh_M

Answer 13

U know Q now so check whether its greater than or lesser than Kp

Answer 14

It is not greater, therefore it is less. so the answer would be C

Answer 15

@Prathamesh_M

Answer 16

if Q<Kp then the reaction proceeds to the right

Answer 17

Remember it like this: the value of Q must adjust itself to equal the value of Kp. if Q increases it means concentration of the products is increasing so the reaction is proceeding to the right and vice versa.