Consider the reaction:

NH4HS(s)ΔNH3( g) + H2S( g)

At a certain temperature, Kc = 8.5 * 10 - 3. A reaction mixture at this temperature containing solid NH4HS has [NH3] = 0.0822 M and [H2S] = 0.0822M. Will more of the solid form, or will some of the existing solid decompose as equilibrium is reached?

Video transcript

Hello everyone. So in this video we're gonna go ahead and take a look at this equation over here and we're just seeing if more of the CCO three will form or will some of the existing C C 03 to compose as equilibrium is reached in this reaction. So some things we need to go ahead and take a look at is first of all we need of course calculate for Q. If Q is equal to K value, then the reaction is at equilibrium. If our Q value is less than r K value, then the reaction shifts to the forward direction. And as you have guessed if Q is greater than R K value, then the reaction will shift in the reverse direction to reach equilibrium. Alright, so to go ahead and calculate for a Q value that's going to equal to the concentration of carbon dioxide. So these brackets just indicates the concentration value. And let's see here the Q value Is going to equal two or 0.215. And why is that? It's because we see that the concentration of CO2 is going to be given to us right there. That's why I wrote this here. So Because this 0.215 value is greater than our constant value which is listed right over here. So let's go ahead and actually highlight these to the problem. So we see that obviously 0.215 is greater than 0.0132 in the case. This just means that the reaction will go ahead and shift to the reverse direction to go ahead and try to reach equilibrium. So formally. Our answer then, is that if the equilibrium will ship in the reverse direction, then more solids will form. So our answer is that more CAC 03 four, And this is going to be my final answer for this problem. Thank you all so much for watching.