Answered

Get the most out of your questions with IDNLearn.com's extensive resources. Get accurate and comprehensive answers to your questions from our community of knowledgeable professionals.

Consider the reaction: [tex] N_2(g) + 3 Br_2(g) \rightleftharpoons 2 NBr_3(g) [/tex]

At equilibrium, the concentrations of [tex] N_2 [/tex] and [tex] Br_2 [/tex] are 0.34 M and 0.70 M respectively, and the concentration of [tex] NBr_3 [/tex] is 0.090 M. What is [tex] K_c [/tex] for this equilibrium?

Sagot :

GinnyAnsweridn
To determine the equilibrium constant [tex]\( K_c \)[/tex] for the given reaction:

[tex]\[ \text{N}_2 (g) + 3 \text{Br}_2 (g) \rightleftharpoons 2 \text{NBr}_3 (g) \][/tex]

we use the equilibrium concentrations of the reactants and products provided:

- The concentration of [tex]\( \text{N}_2 \)[/tex] is 0.34 M.
- The concentration of [tex]\( \text{Br}_2 \)[/tex] is 0.70 M.
- The concentration of [tex]\( \text{NBr}_3 \)[/tex] is 0.090 M.

The expression for the equilibrium constant [tex]\( K_c \)[/tex] for this reaction is given by:

[tex]\[ K_c = \frac{[\text{NBr}_3]^2}{[\text{N}_2] \cdot [\text{Br}_2]^3} \][/tex]

Now, we can substitute the given concentrations into this expression:

[tex]\[ K_c = \frac{(0.090)^2}{(0.34) \cdot (0.70)^3} \][/tex]

First, calculate the numerator:

[tex]\[ (0.090)^2 = 0.090 \times 0.090 = 0.0081 \][/tex]

Next, calculate the denominator:

[tex]\[ (0.70)^3 = 0.70 \times 0.70 \times 0.70 = 0.343 \][/tex]
[tex]\[ (0.34) \times (0.343) = 0.11662 \][/tex]

Therefore:

[tex]\[ K_c = \frac{0.0081}{0.11662} \approx 0.0695 \][/tex]

So, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction is approximately 0.0695.
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.