## 5J.13

Tracy Tolentino_2E
Posts: 140
Joined: Sat Sep 07, 2019 12:17 am

### 5J.13

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=1.7x10^-3 at 600.K and 7.8x10^-5 at 700. K.

How do you solve this?

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### Re: 5J.13

Since the K value at 700K is smaller, it means that more reactants are formed since K = Products/Reactants. Since there are more reactants, it means that at 700K, there will be less products and so less ammonia will be present at the higher temperature of 700K.

Tracy Tolentino_2E
Posts: 140
Joined: Sat Sep 07, 2019 12:17 am

### Re: 5J.13

Do we not have to calculate anything?

Simon Dionson 4I
Posts: 107
Joined: Sat Sep 14, 2019 12:17 am

### Re: 5J.13

Tracy Tolentino_2E wrote:Do we not have to calculate anything?

I didn't do any math, I used the K values to base my answers. Since K at the lower temperature was smaller, I assumed more products were formed at 600K vs 700K.

MAC 4G
Posts: 121
Joined: Wed Sep 18, 2019 12:16 am

### Re: 5J.13

Tracy Tolentino_2E wrote:Do we not have to calculate anything?

This question doesn't require you to calculate anything, it just wants you to take the given information and see if you understand that if K at 700K is smaller, then it would lead you to the conclusion that there are more reactant present. Therefore, there would be less ammonia present at the higher temperature, assuming there wasn't any other changes to the system.

Return to “Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions”

### Who is online

Users browsing this forum: No registered users and 2 guests