Answer
$$K_{c} = 1.60 \times 10^{57}$$
Work Step by Step
1. The reaction was reversed. Therefore:
$$K_{c2} = 1/(K_{c1}) = 1/(6.25 \times 10^{-58}) = 1.60 \times 10^{57}$$
Chemistry: The Molecular Science (5th Edition)
by Moore, John W.; Stanitski, Conrad L.
Published by
Cengage Learning
ISBN 10:
1285199049
ISBN 13:
978-1-28519-904-7
Chapter 12 - Chemical Equilibrium - Exercise 12.4 - Manipulating Equilibrium Constants - Page 532: b
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