At 745K, the reaction 2AB(g) ? A2(g) + B2(g) has a Kc = 2.00 x 10 Solution React

Question

At 745K, the reaction 2AB(g) ? A2(g) + B2(g) has a Kc = 2.00 x 10

Explanation / Answer

Reaction equation 2 AB(g) ? A2(g) + B2(g) Initial concentrations of AB and A2 are [AB]0 = [A2]0 = 1 mol / 10.0 L = 0.1 M Equilibrium equation Kc = [A2]·[B2]/[AB]² with Kc = 0.02 ICE table .......... [AB].......... [A2]........... [B2] I.......... 0.1............ 0.1............ 0 C........ -2·x........... +x............. +x E...... 0.1 - 2·x.... 0.1 + x........... x Hence, 0.02 = (0.1 + x)·x/(0.1 - 2·x)² 0.02·(0.01 - 0.4·x + 4·x²) = 0.1·x + x² 0.92·x² + 0.108·x - 0.0002 = 0 => x = ( -0.108 ±v(0.108² + 0.0002·4·0.92) ) / ( 2·0.92 ) => x = -0.119 M x = 0.00183 M First solution is infeasible because it would lead to a negative concentrations for A2 and B2 So the equilibrium concentration of B2 is [B2] = x = 0.00183 MReaction equation 2 AB(g) ? A2(g) + B2(g) Initial concentrations of AB and A2 are [AB]0 = [A2]0 = 1 mol / 10.0 L = 0.1 M Equilibrium equation Kc = [A2]·[B2]/[AB]² with Kc = 0.02 ICE table .......... [AB].......... [A2]........... [B2] I.......... 0.1............ 0.1............ 0 C........ -2·x........... +x............. +x E...... 0.1 - 2·x.... 0.1 + x........... x Hence, 0.02 = (0.1 + x)·x/(0.1 - 2·x)² 0.02·(0.01 - 0.4·x + 4·x²) = 0.1·x + x² 0.92·x² + 0.108·x - 0.0002 = 0 => x = ( -0.108 ±v(0.108² + 0.0002·4·0.92) ) / ( 2·0.92 ) => x = -0.119 M x = 0.00183 M First solution is infeasible because it would lead to a negative concentrations for A2 and B2 So the equilibrium concentration of B2 is [B2] = x = 0.00183 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.