Consider the following system at equilibrium where AH\' -198 kJ, and Ke-2.90-10
ID: 1038593 • Letter: C
Question
Consider the following system at equilibrium where AH' -198 kJ, and Ke-2.90-10 If the VOLUME of the equlibrium system is sudenly decreased at constant temperature: The value of Ke A increases B. decreases C. remains the same. The value of Qe A. is greater than Kc B. is equal to Ke C. is less than Ke. The reaction must: A. run in the forward direction to reestablish equilibrium. B. run in the reverse direction to reestablish equilibrium remain the same. It is already at equilibrium The number of moles of O, wialA increa A increase B. decrease C. remain the same.Explanation / Answer
The equilibrium constant will not change with change in volume or concentration of reactants or product. So the value of Kc will remains same.
The reaction is in equilibrium, so Q = Kc or reaction quotient is equal to equilibrium constant.
If the volume decreased, according to Lechatelier principle, the system will try to nullify its effect. That is volume decreased means pressure increased, so the system will try to decrease the pressure by favoring the reaction in which number of moles decreasing , that is backward reaction will be more faster in which 3 moles of product (2SO2 + 1O2 ) becomes to 2 moles of reactant (2SO3). Run in the reverse direction to reestablish equilibrium.
So the number of moles of O2 will also decrease, because it is in product side.