1.

What is the value of equilibrium constant based on concentrations if the forward and reverse rate constant of a non – equilibrium reaction are 1.85 × 10^-4 and 2.34 × 10^-5?(a) 6.45(b) 7.91(c) 8.43(d) 2.31I got this question during an interview.My doubt is from Chemical Equilibrium in High Temperature Air in section Properties of High Temperature Gases of Aerodynamics

Answer»

The correct ANSWER is (b) 7.91

To elaborate: Given, KF = 1.85 × 10^-4 , kb= 2.34 × 10^-5

The EQUILIBRIUM constant based on concentrations for non – equilibrium is same as derived for the equilibrium condition SINCE it relates the forward and reverse rate constants. It is given by:

kc = \(\frac {k_f}{k_b}\)

Substituting the values, we get,

kc = \(\frac {1.85 × 10^{- 4}}{2.34 × 10^{- 5}}\) = 7.91



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