For the decomposition of SO₃(g),K_c = [SO₂]²[O₂]/[SO₃]²,at equilibrium,there are 0.090 mol SO₂,0.110 mol O₂,0.100 mol SO₃ in a 25.0-L container.What is the value of K_c?

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Quizplus · Accepted Answer

The equilibrium constant $K_c$ is calculated using the concentrations of the products and reactants at equilibrium. Given the moles of each substance and the volume of the container (25.0 L), we can find the concentrations:- $[SOS] = \frac{0.100 \, \text{mol}}{25.0 \, \text{L}} = 0.0040 \, \text{M}$- $[SO_2] = \frac{0.090 \, \text{mol}}{25.0 \, \text{L}} = 0.0036 \, \text{M}$- $[O_2] = \frac{0.110 \, \text{mol}}{25.0 \, \text{L}} = 0.0044 \, \text{M}$Plugging these into the equilibrium expression:$$K_c = \frac{[SO_2][O_2]}{[SOS]} = \frac{(0.0036)(0.0044)}{0.0040} = 3.96 \times 10^{-3}$$Rounding to two significant figures gives $3.6 \times 10^{-3}$, which matches choice A.

For the Decomposition of SO3(g),Kc = [SO2]2[O2]/[SO3]2,at Equilibrium,there Are 0

For the Decomposition of SO₃(g),K_c = [SO₂]²[O₂]/[SO₃]²,at Equilibrium,there Are 0