5J.13 textbook problem

[Varsha Ravi 3E]

I'm having a little trouble figuring out how to set up this problem. Does anyone know how to solve it? A gaseous mixture consisting of 2.23 mmol N2 and 6.69 mmol H2 in a 500.-mL container was heated to 600. K and allowed to reach equilibrium. Will more ammonia be formed if that equilibrium mixture is then heated to 700. K? For N2(g) + 3H2(g) > 2NH3(g), K = 1.7 x 10^-3 at 600. K and 7.8 x 10^-5 at 700. K

[Yu Jin Kwon 3L]

Hi! So this is kinda unlike our other problems where we did ICE box because they give you the K at the different temperatures, meaning we can just directly compare the different K values at the different temperatures. In this case, you want to find which one is the larger K because that would mean it favors products (aka ammonia) more (which is what we want). So, if we compare K = 1.7 x 10^-3 at 600K and K = 7.8 x 10^-5 at 700K, we can see that the K at 600K is bigger, SO to answer that question, less ammonia will be formed at 700K.

Hope this helps!

[Samantha Pedersen 2K]

For this problem, you actually don't need to set up or solve any calculations. Notice that the value of K is smaller at 700. Kelvin than at 600. Kelvin. This means that there must be a smaller number in the numerator (where the equilibrium concentrations of the products go) of the equilibrium constant expression at 700. Kelvin in order to get a smaller value for K. Ammonia is the product of the reaction, so we know that a smaller amount of ammonia will be formed if the equilibrium mixture is heated to 700. Kelvin. I hope this helps!

[Varsha Ravi 3E]

Hi! So this is kinda unlike our other problems where we did ICE box because they give you the K at the different temperatures, meaning we can just directly compare the different K values at the different temperatures. In this case, you want to find which one is the larger K because that would mean it favors products (aka ammonia) more (which is what we want). So, if we compare K = 1.7 x 10^-3 at 600K and K = 7.8 x 10^-5 at 700K, we can see that the K at 600K is bigger, SO to answer that question, less ammonia will be formed at 700K.

Hope this helps!

Thank you so much!

[Silvi_Lybbert_3A]

As an added note, if you did want to do calculations to confirm such conceptual reasoning, I think this is a good example of when the van't Hoff equation would be acceptable to use. If you plug in the values given to the equation and solve for the standard change in enthalpy, you will get a negative value. This implies this equation is exothermic and, if heat is added, the reverse endothermic process will be favored. I am not 100% sure van't Hoff equation is acceptable here, so maybe someone can confirm.