Learning Goal: To understand the relationship between the equilibrium constant a

Question

Learning Goal: To understand the relationship between the equilibrium constant and rate constants. For a general chemical equation A+BC+D the equilibrium constant can be expressed as a ratio of the concentrations: Kc=[C][D][A][B] If this is an elementary chemical reaction, then there is a single forward rate and a single reverse rate for this reaction, which can be written as follows: forward ratereverse rate==kf[A][B]kr[C][D] where kf and kr are the forward and reverse rate constants, respectively. When equilibrium is reached, the forward and reverse rates are equal: kf[A][B]=kr[C][D] Thus, the rate constants are related to the equilibrium constant in the following manner: Kc=kfkr=[C][D][A][B]

Part A For a certain reaction, Kc= 7.63×106 and kf= 1.82×104 M2s1 . Calculate the value of the reverse rate constant, kr. Express your answer numerically in inverse seconds. Hints kr = m M2s1 SubmitMy AnswersGive Up

Part B For a different reaction, Kc=4.22×107, kf=3.85×104s1, and kr= 9.12×104 s1 . Adding a catalyst increases the forward rate constant to 1.17×107 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst? Express your answer numerically in inverse seconds. Hints kr = m s1 SubmitMy AnswersGive Up

Part C Yet another reaction has an equilibrium constant Kc=4.32×105 at 25 C. It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 C , what will happen to the equilibrium constant? Hints The equilibrium constant will Yet another reaction has an equilibrium constant at 25 . It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 , what will happen to the equilibrium constant? increase. decrease. not change. SubmitMy AnswersGive Up

Explanation / Answer

Kr = kc / k f=(7.63*10^6)/1.82*10^4= 419.230 M^-2s^-1

(B) new kr = kc/kf = (4.22*10^7)/(1.17*10^7)=3.60s^-1

(C)the reaction will shift towards the reactant side, therefore equilibrium constant should decrease.

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

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