Question

A solution is made by mixing 30.0 mL of 0.150M (mol/L) compound A with 25.0 mL of 0.200M (mol/L) compound B. At equilibrium, the concentration of C is 0.0454M. Calculate the equilibrium constant, K, for this reaction.

Answer 1

Answer:

Keq = 6.105

Explanation:

• A + B → C

∴ Keq = [ C ] / ( [ A ] * [ B ] )

∴ [ C ] = 0.0454 M

⇒ [ A ] = ((0.030 L)(0.150 M)) / (0.03 + 0.025) = 0.0818 M

⇒ [ B ] = ((0.025 L)(0.200)) / (0.03 + 0.025) = 0.0909 M

⇒ Keq = ( 0.0454) / ((0.0818)(0.0909))

⇒ Keq = 6.105

Answer 2

The equilibrium constant, K is $6.105$

Given that, A solution is made by mixing 30.0 mL of 0.150M (mol/L) compound A with 25.0 mL of 0.200M (mol/L) compound B.

Reaction is,   A + B ⇒ C

The equilibrium constant K is given as,

                                   $K=\frac{[C]}{[A]*[B]}$

$[C]=0.0454M$

$[A]=\frac{0.030*0.150}{0.03+0.025}=0.0818M$

$[B]=\frac{0.025*0.200}{0.03+0.025}=0.0909M$

Now calculate equilibrium constant,

              $K_{eq.}=\frac{0.0454}{(0.0818)*(0.0909)}=6.105$

Hence,  the equilibrium constant, K is $6.105$

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