Number 2 please and thank you;) , A solution of 5.00 mL of 0.00300 M Fe(NO,)h,4.

Question

Number 2 please and thank you;)

, A solution of 5.00 mL of 0.00300 M Fe(NO,)h,4.00 ml of 0.00300 M KSCN, and 3.00 mL of 10 M HNO, is mixed together and allowed to reach equilibrium. The concentration of Fe(SCN) is found to be 2.72x10 M at equilibrium. Calculate the equilibrium constant for this reaction. 3. Given the following reaction: A(g) +2B(g) 3C(8) 4.00 moles of A and 3.00 moles of B are combined in a 2.0 L container. After equilibrium is obtained, it is found that 0.038 mol/L of B are present. Calculate the equilibrium constant of this reaction.

Explanation / Answer

Ans 2

The balanced reaction

Fe3+(aq) + SCN-(aq) -----> Fe(SCN)2+ (aq)

Moles of Fe(NO3)3 = molarity x volume

= 0.00300 Mol/L x 5 mL x 1L/1000mL

= 0.000015 mol

Moles of KSCN = molarity x volume

= 0.00300 M x 4 mL x 1L/1000 mL

= 0.000012 mol

total volume of solution = 5 + 4 + 3 = 12 mL x 1L/1000 mL

= 0.012 L

After mixing

Concentration of [SCN-] = moles of KSCN / total volume

= 0.000012 mol / 0.012 L

= 0.00100 M

Concentration of [Fe3+] = moles of Fe(NO3)3 / total volume

= 0.000015 mol / 0.012 L

= 0.00125 mol

at equilibrium

Concentration of [Fe3+] = 0.00125-x

Concentration of [SCN-]= 0.00100 -x

At equilibrium, concentration of Fe(SCN)2+ = 2.72 x 10^-4 M

x = 2.72 x 10^-4

Concentration of [Fe3+] = 0.00125-( 2.72 x 10^-4)

= 0.000978 M

Concentration of [SCN-]= 0.00100 -( 2.72 x 10^-4)

= 0.000728 M

Equilibrium constant expression of the

K = [Fe(SCN)2+] / [Fe3+] [SCN-]

K = (2.72 x 10^-4) /( 0.000978)*( 0.000728)

= 382.03

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