Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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]
[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]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
I need help with piecewise functions. How to find the domain and range and how to write it notation.