Question

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1. To find Keq for the reaction :

Fe3+ (aq) + SCN- (aq) ↔ FeSCN2+ (aq)

A student mixes 5.0 mL of 1.00x10-3 M Fe(NO3)3 with 5.0 mL 1.00x10-3 M KSCN and finds that in the equilibrium mixture the concentration of FeSCN2+ is 1.0x10-4 M.

Find the number of moles Fe3+ and SCNinitially present.

__________ moles Fe3+; __________ moles SCN

How many moles of FeSCN2+ are in the mixture at equilibrium?

__________ moles FeSCN2+

How many moles of Fe3+ and SCNare used up in making the FeSCN2+?

__________ moles Fe3+; __________ moles SCN

How many moles of Fe3+ and SCNremain in the solution at equilibrium?

__________ moles Fe3+; __________ moles SCN

What are the concentrations of Fe3+, SCN- , and FeSCN2+ at equilibrium?

[Fe3+] = __________ M; [SCN- ] = __________ M; [FeSCN2+] = __________ M

What is the value of Keq for the reaction?

Keq = __________

Answer 1

1) The moles of the ions are calculated:

n Fe + 3 = M * V = 1x10 ^ -3 M * 0.005 L = 5x10 ^ -6 mol

n SCN- = 1x10 ^ -3 M * 0.005 L = 5x10 ^ -6 mol

2) The moles of the product in the equilibrium are calculated:

n FeSCN + 2 = 1x10 ^ -4 * 0.01 = 1x10 ^ -6 mol

3) 1 mole of FeSCN + 2 is formed for each mole of each reagent, then:

n Fe + 3 used = n SCN- used = 1x10 ^ -6 mol

4) The remaining moles of each reagent are calculated:

n Fe + 3 remaining = n SCN- remaining = 5x10 ^ -6 - 1x10 ^ -6 = 4x10 ^ -6 mol

5) The equilibrium concentrations are calculated:

[Fe + 3] = n / V = ​​4x10 ^ -6 / 0.01 = 0.0004 M

[SCN-] = 4x10 ^ -6 / 0.01 = 0.0004 M

[FeSCN + 2] = 1x10 ^ -6 / 0.01 = 0.0001 M

6) Keq is calculated:

Keq = [FeSCN + 2] / [Fe + 3] * [SCN-] = 0.0001 / (0.0004) ^ 2 = 625

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