Saved Bookmarks
| 1. |
Consider the following reaction at \( 600^{\circ} C \). \[ 2 SO _{2}( g )+ O _{2}( g ) \rightleftharpoons 2 SO _{3}( g ) ; \quad K_{C}=4.00 \] A mixture was prepared with \( \left[ SO _{3}\right]=0.50 M ,\left[ O _{2}\right]=0.35 M \) and \( \left[ SO _{2}\right]=0.0 M \) Thus, equilibrium concentration of \( O _{2} \) is (a) \( 0.0450 M \) (b) \( 0.035 M \) (c) \( 0.35 M \) (d) \( 0.45 M \) |
||||||||
|
Answer» Correct option is (d) 0.45 M Given, \(2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)\) \(K_C = \frac{[SO_3]^2}{[SO_2]^2[O_2]^2}\)
\(\therefore K_C= \frac{(0.5 + 2x)^2}{(-2x)^2(0.35 -x)}\) ⇒ \(4 = \frac{0.25 + 4x^2 + 2x}{4x^2 (0.35 -x)}\) ⇒ \(16x^2 (0.35 - x) = 0.25 + 4x^2+ 2x\) ⇒ \(5.6 x^2 - 16x^3= 0.25+4x^2+ 2x\) ⇒ \(16 x^3 - 1.6 x^2 + 2x + 0.25 = 0\) ⇒ \(x = - 0.10633\) ∴ Concentration of O2 at equilibrium \(= 0.35 - (-0.10633)\) \(= 0.35 + 0.10633\) \(= 0.456 M\) |
|||||||||