Search OWv2|Online teaching a...x Use the Refereaces to access important values

Question

Search OWv2|Online teaching a...x Use the Refereaces to access important values if needed for this question te following system at equilbrium where .10.4 k, and K-556, at 698 K When 0.24 moles of Ia() are removed from the eqailbbrium system at constant temperatare the value ofK,A increases. B decreases C remains the same the value of Q'[ Ausgreater than K. B. is equal to K C. is less than K the reaction mmst: A rua mthe forward direction to reestablish equitran B. run in the reverse direction to reestablish eqlibim C. remain the same. It is already at equilabrium the concentration of H, will:A increase B decrease C. remain the same

Explanation / Answer

The value of Kc remains the same

The value of Qc is greater than Kc

Qc = products / Reactants

Qc = [2HI]2 / ([H2] * [I2] ), if the concentration of one of the reactants decreases the value of Qc increases

The reaction must meet lecheteliers principle, the system must reach an equilibrium after the removal of I2, so the reaction must run in the reverse direction

The concentration of H2 according to the last answer will increase

*please rate the answer if you like it =)

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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