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Sauron [17]
3 years ago
12

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 equilibriu

m is established: 2NO(g)+2H2(g)⥫⥬==N2(g)+2H2O(g) At equilibrium [NO]=0.062M. Calculate the equilibrium concentrations of H2, N2, and H2O.
Calculate the equilibrium concentration of N2
Chemistry
1 answer:
tatyana61 [14]3 years ago
5 0

Answer:

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

                  2NO(g)         +    2H2(g)    ⇒        N2(g)      +         2H2O(g)

i  mol            0.10                   0.050                                             0.10

c mol            -0.038                -0.038                +0019                +0.038                                                

e mol            0.062                 0.012                  00.019               0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] =   0.062 M

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

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Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You
Sliva [168]

Answer:

0.11%

Explanation:

Without mincing words, let us dive straight into the solution to the question/problem. The first step to solve this question is to write out the chemical reaction, that is the reaction showing the dissociation of acetic acid.

CH3COOH <=======================================> CH3COO⁻ + H⁺

Initially, the amount present in the acetic acid which is = 12M, the concentration for CH3COO⁻  and H⁺ is 0 respectively.

At equilibrium, the amount present in the acetic acid which is = 12 - x, the concentration for CH3COO⁻ = x  and H⁺ = x respectively. Note that the ka for acetic acid = 1.8 × 10⁻⁵.

1.8 × 10⁻⁵ = x²/ 14 - x. Therefore, x = 0.0158 M.

The next thing to do is to calculate for the percentage of dissociation, this can be done as given below:

percentage of dissociation = x/14 × 100. Recall that the value that we got for x = 0.0158 M. Hence, the percentage of dissociation = 0.0158 M/ 14m × 100 = 0.11%

7 0
2 years ago
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