Question

question 26 of 32
the table below gives the equilibrium concentrations for this reaction at a certain temperature:
n₂(g) + o₂(g) ⇌ 2no(g)

n₂	o₂	no
0.69 m	0.98 m	0.034 m

what is the equilibrium constant for the reaction?
a. 9.9
b. 20
c. 5.0×10⁻²
d. 1.7×10⁻³

Explanation:

Step1: Recall the equilibrium constant formula

For the reaction \( \ce{N2(g) + O2(g)
ightleftharpoons 2NO(g)} \), the equilibrium constant \( K_c \) is given by the formula \( K_c=\frac{[\ce{NO}]^2}{[\ce{N2}][\ce{O2}]} \).

Step2: Substitute the given concentrations

We know that \( [\ce{N2}] = 0.69\ M \), \( [\ce{O2}] = 0.98\ M \), and \( [\ce{NO}] = 0.034\ M \). Substituting these values into the formula:
\[
K_c=\frac{(0.034)^2}{(0.69)(0.98)}
\]

Step3: Calculate the numerator and the denominator

First, calculate the numerator: \( (0.034)^2 = 0.034\times0.034 = 0.001156 \)
Then, calculate the denominator: \( (0.69)(0.98)=0.6762 \)

Step4: Divide the numerator by the denominator

\[
K_c=\frac{0.001156}{0.6762}\approx 1.7\times 10^{-3}
\]

Answer:

D. \( 1.7\times 10^{-3} \)