Consider the equilibrium shown below. When one looks at the equilibrium where X = CH 3 and the one where X = CH(CH 3 ) 2 , how do the values of the equilibrium constants (Ks) compare? 3 and the one where X = CH(CH 3 ) 2 , how do the values of the equilibrium constants (Ks) compare? A) The Ks are equal. B) The K where X = CH 3 is greater. C) The K where X = CH(CH 3 ) 2 is greater. D) The Ks differ only slightly and are both less than 1. class=answers-bank-image d-block rel=preload A) The Ks are equal.B) The K where X = CH 3 is greater.C) The K where X = CH(CH 3 ) 2 is greater.D) The Ks differ only slightly and are both less than 1.

Question

Consider the equilibrium shown below. When one looks at the equilibrium where X = CH 3  and the one where X = CH(CH 3 )  2 , how do the values of the equilibrium constants (Ks)  compare?  3  and the one where X = CH(CH 3 )  2 , how do the values of the equilibrium constants (Ks)  compare?   A)  The Ks are equal. B)  The K where X = CH 3  is greater. C)  The K where X = CH(CH 3 )  2  is greater. D)  The Ks differ only slightly and are both less than 1. class=answers-bank-image d-block rel=preload >A)  The Ks are equal.B)  The K where X = CH 3  is greater.C)  The K where X = CH(CH 3 )  2  is greater.D)  The Ks differ only slightly and are both less than 1.

Garcia Baldwin · Accepted Answer

Final Answer : C, Explanation :  The addition of the CH group to the peptide sequence increases the complexity of the molecule, making it less likely to dissociate into its constituent parts, which would decrease the value of the equilibrium constant. Therefore, the equilibrium where X = CH(CH 3 ) 2 , which has the additional CH group, will have a higher K value compared to where X = CH 3 .

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