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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 15 - Problem 91p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 15 - Problem 91p

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# Consider the reaction. A solution is made containing initial [Fe3+] = 1.0 × 10?3 M and

ISBN: 9780321910295 34

## Solution for problem 91P Chapter 15

Introductory Chemistry | 5th Edition

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Problem 91P

Consider the reaction.

A solution is made containing initial [Fe3+] = 1.0 * 10-3 M and initial [SCN-] = 8.0 * 10-4 M. At equilibrium, [FeSCN2+] = 1.7 * 10-4 M. Calculate the value of the equilibrium constant. Hint: Use the chemical reaction stoichiometry to calculate the equilibrium concentrations of Fe3+ and SCN-.

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the value of the equilibrium constant.

Step1:

The equilibrium constant for a general reaction,

aA + bB ⇄ cC + dD

is expressed as,

Kc = [C]c [D]d / [A]a [B]b

Where [A], [B], [C] and [D] are the equilibrium concentrations of the reactants and products.

Step2:

The given reaction is:

Fe3+(aq) + SCN-(aq) ⇄ FeSCN2+(aq)

Let ‘x’ be the concentration of Fe3+ and SCN- reacted to form ‘x’ FeSCN2+

Given,

Initial concentration of Fe3+ = 1.0 x 10-3 M

Initial concentration of SCN- = 8.0 x 10-4 M

Equilibrium concentration of FeSCN2+ = 1.7 x 10-4 M

This indicates that amount of Fe3+ and SCN- reacted = 1.7 x 10-4 M

Therefore, equilibrium concentration of Fe3+ = (1.0 x 10-3 M) - (1.7 x 10-4 M)

= 8.3 x 10-4 M

Equilibrium concentration of SCN- = (8.0 x 10-4 M) - (1.7 x 10-4 M)

= 6.3 x 10-4 M

Step3:

The equilibrium concentration, Kc for the above reaction is:

Kc = [FeSCN2+] / [Fe3+] [SCN-]

= (1.7 x 10-4 M) / [( 8.3 x 10-4 M) (6.3 x 10-4 M)]

= (1.7 x 10-4 M) / (5.229 x 10-7)

= 0.3251 x 103

= 325.1

Thus, the value of equilibrium constant is 325.1.

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Step 2 of 3

Step 3 of 3

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Consider the reaction. A solution is made containing initial [Fe3+] = 1.0 × 10?3 M and