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A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O...

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established: 2NO(g)+2H2(g)←−→N2(g)+2H2O(g) At equilibrium [NO]=0.062M. Calculate the equilibrium concentration of H2. Calculate the equilibrium concentration of N2. Calculate the equilibrium concentration of H2O. Calculate Kc

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Expert Solution

When an excercise gives you data about initial amounts on a vessel you have to check the direction of the reaction. Sometimes the reaction goes from products to reagents is not always to the right. In order to check the direction look the data, if you pay attention, data of a product is missing, so the reaction must go to the right in order to produce this product. Also, the concentration of the reagent NO at the equilibrium is lower than the initial, that means that the reaction goes to the right.

Why i tell you this? because you have to know if the concentrations increase or decrease to do an ICE table:

2NO + 2H2 <------> N2 + 2H2O

0.1____0.05_____0____0.1

0.1-2x__0.05-2x__x____0.1+2x

The reagents are being consumed so i have to put -2x on the ICE table, if the reaction is backwards the reagents are being formed so i have to put +2x.

Now, let`s continue with the excercise:

We know [NO]=0.062M= 0.1-2x ----->x= 0.019M

Kc= [N2][H2O]2/[NO]2[H2]2

[NO]= 0.062M

[H2]= 0.05-2x= 0.012M

[N2]= 0.019M

[H2O]= 0.1 + 2x= 0.138M

Kc= (0.019)(0.138)2/(0.062)2(0.012)2= 653.7

queen_honey_blossom answered 2 years ago
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