Question

1. if the equilibrium concentrations found in the reaction:

a(g) + 2 b(g) ⇌ 2 c(g)
are a = 0.025 m, b = 0.15 m, and c = 0.55 m, calculate the
value of kc.

Explanation:

Step1: Write the equilibrium constant expression

For the reaction \(A(g)+2B(g)
ightleftharpoons 2C(g)\), the equilibrium constant \(K_{c}\) is given by \(K_{c}=\frac{[C]^{2}}{[A][B]^{2}}\)

Step2: Substitute the given concentrations

Substitute \([A]=0.025M\), \([B] = 0.15M\), and \([C]=0.55M\) into the expression:

$$
LATEXBLOCK0
$$

Step3: Calculate the value of \(K_{c}\)

$$K_{c}\approx538$$

Answer:

\(K_{c}\approx538\)