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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.
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