Consider the following equilibrium for which Delta H = -75.54 kJ: 2 NO(g) + Cl_2
ID: 483296 • Letter: C
Question
Consider the following equilibrium for which Delta H = -75.54 kJ: 2 NO(g) + Cl_2(g) 2 NOCl(g) If a mixture of the gases Involved In this reaction is allowed to come to equilibrium, and the reaction mixture is cooled, what will the reaction do in response to this change, and how will the value of K (the equilibrium constant) be affected? The reaction will not shift and the value of K will not change. The reaction will shift toward reactants with a decrease in K. The reaction will shift toward reactants with an increase In K. The reaction will shift toward reactants with no change in K. The reaction will shift toward products with an Increase in K. Consider the following equilibrium: 2 H_2O(l) + 2 Cl_2(g) 4 HCI(g) + O_2(g) If a mixture of the chemicals Involved in this reaction is allowed to come to equilibrium, and some H_2O is removed from the reaction mixture, what will the reaction do in response to this change, and how will the value of K (the equilibrium constant) be affected? The reaction will shift toward reactants with no change in K. The reaction will shift toward reactants with a decrease in K. The reaction will shift toward products with a decrease in K. The reaction will not shift and the value of K will not change. The reaction will shift toward products with an increase in K.Explanation / Answer
Ans for 13:
according to vant hoff equation
ie ln (K2/K1)= -delta H/( 1/T2-1/T1)
in this reaction as temperature is deceased due to cooling, T2<T1.
hence K2<K1.
So option (b)
Ans for 14:
According to Le-Chatelier's principle
when stress is applied on equilibrium system, then it adjust itself by moving in forward or backward direction in order to nullify the stress applied.
in the given equilibrium if concentration of reactants are increased then reaction will shifted in forward direction ie towards product side and vece versa. This is application of Le chateliers principle.
as H2O is amongst the reactants, hence when it is removed from reaction then to nullify this stress the reaction will be shifted towards reactant side to achieve the equilibrium back.
since equilibrium is reformed again the equilibruim constant will not change.
so answer option (a).