emilyahope39
Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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 :

GinnyAnswer
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]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.