Emilyahope39idn
Answered

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Consider the following reaction:

[tex]\[2 NO (g) + Br_2 (g) \rightleftharpoons 2 NOBr (g)\][/tex]

At a specific temperature, the equilibrium concentrations were determined to be [tex]\([NO] = 0.089\, M\)[/tex], [tex]\([Br_2] = 0.070\, M\)[/tex], and [tex]\([NOBr] = 0.183\, M\)[/tex].

(a) What is the value of the equilibrium constant?

[tex]\[K_{eq} = \square\][/tex]

(b) Describe the position of the equilibrium.

The reaction is (select) [tex]\(\square\)[/tex]

Sagot :

GinnyAnsweridn
Given the reaction:
[tex]\[ 2 \text{NO} (g) + \text{Br}_2 (g) \rightleftharpoons 2 \text{NOBr} (g) \][/tex]

and the equilibrium concentrations:
[tex]\[ [\text{NO}] = 0.089 \, M \][/tex]
[tex]\[ [\text{Br}_2] = 0.070 \, M \][/tex]
[tex]\[ [\text{NOBr}] = 0.183 \, M \][/tex]

We need to determine:
(a) The value of the equilibrium constant, [tex]\( K_{eq} \)[/tex].
(b) The position of the equilibrium.

### Step by Step Solution:

#### (a) Calculating the Equilibrium Constant, [tex]\( K_{eq} \)[/tex]:

1. Write the expression for the equilibrium constant, [tex]\( K_{eq} \)[/tex], for the given reaction:
[tex]\[ 2 \text{NO} (g) + \text{Br}_2 (g) \rightleftharpoons 2 \text{NOBr} (g) \][/tex]
The equilibrium constant expression is:
[tex]\[ K_{eq} = \frac{[\text{NOBr}]^2}{[\text{NO}]^2 [\text{Br}_2]} \][/tex]

2. Substitute the equilibrium concentrations into the expression:
[tex]\[ [\text{NOBr}] = 0.183 \, M \][/tex]
[tex]\[ [\text{NO}] = 0.089 \, M \][/tex]
[tex]\[ [\text{Br}_2] = 0.070 \, M \][/tex]

3. Calculate the equilibrium constant:
[tex]\[ K_{eq} = \frac{(0.183)^2}{(0.089)^2 \times 0.070} \][/tex]

After performing the calculations, we find:
[tex]\[ K_{eq} \approx 60.398 \][/tex]

Thus,
[tex]\[ K_{eq} = 60.398 \][/tex]

#### (b) Determining the Position of the Equilibrium:

To determine the position of the equilibrium, we compare the value of [tex]\( K_{eq} \)[/tex] to 1:

- If [tex]\( K_{eq} > 1 \)[/tex], the reaction favors the formation of products.
- If [tex]\( K_{eq} < 1 \)[/tex], the reaction favors the formation of reactants.
- If [tex]\( K_{eq} \approx 1 \)[/tex], the reaction has no strong preference for either reactants or products.

Given that [tex]\( K_{eq} = 60.398 \)[/tex]:

Since [tex]\( K_{eq} \)[/tex] is significantly greater than 1, the reaction favors the formation of products.

Therefore, the position of the equilibrium is such that the reaction favors the products.

### To Summarize:

(a) The value of the equilibrium constant is:
[tex]\[ \boxed{60.398} \][/tex]

(b) The reaction is (select):
[tex]\[ \boxed{\text{products}} \][/tex]
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