Question

At 25°C, K = 0.090 for the following reaction. H20(g) Cl20(g)2 HOCI(g) Calculate the concentrations of all species at equilibrium for each of the following cases. (a) 1.3 g H20 and 2.0 g Cl20 are mixed in a 1.1-L flask. [Cl20] [H2O] 四 (b) 1.5 mol pure HOCI is placed in a 1.7-L flask Cl20] [H2O]

Answer 1

(a)

1.3 g H2O (molar mass 18 g/mol) =

(a) 1.3g H_2 O(molar mass 18 g/mol) = 1.3g/18 g/mol = 0.0722 mol 2.0g Cl_2O(molar mass 87 g/mol) 2g/87g/mol = 0.02299 mol x moles of H_2O will react with x moles of Cl_2O to form 2x moles of HOCl The equilibrium concentrations are [H_2O] = 0.0722 - x mol/1.1 L [Cl_2O] = 0.02299 - x mol/1.1 L [HOCl] = 2x mol/1.1 L K = [HOCl]/[H_2O][Cl_2O] 0.090 = [2x mol/1.1 L]^2 /0.0722 - x mol/1.1 L times 0.02299 - x mol/1.1 L 0.090 = 4x^2/(0.0722 - x) times (0.02299 - x) (0.0722 - x) times (0.02299 - x) = 4x^2/0.090 (0.0722 -x) times (0.02299 - x) = 44.44x^2 0.0722(0.02299 - x) - x(0.02299 - x) = 44.44x^2 0.00166 - 0.0722x - 0.02299x + x^2 = 44.44x62 43.44x^2 + 0.095211x - 0.00166 = 0 This is quadratic equation with solution x = -b plusminus squareroot b^2 - 4ac/2a x = -0.095211 plusminus squareroot (0.095211)^2 - 4(43.44)(-0.00166)/2(43.44) x = -0.095211 plusmius 0.0545/86.80 x = 0.00518 or x = -0.00737 The negative value is discarded since the number of moles cannot be negative.

2.0 g Cl2O (molar mass 87 g/mol) $\dfrac { \text { 2 g } }{ \text { 87 g/mol } } = \text { 0.02299 mol }$

x moles of H2O will react with x moles of Cl2O to form 2x moles of HOCl.

The equilibrium concentrations are

$[H_2O] = \frac{ \text { 0.0722 -x mol } }{ \text { 1.1 L } }$

$[Cl_2O] = \frac{ \text { 0.02299 -x mol } }{ \text { 1.1 L } }$

$[HOCl] = \frac{ \text { 2x mol } }{ \text { 1.1 L } }$

$K = \frac{[HOCl]^2}{[H_2O][Cl_2O]}$

$0.090 = \frac{[ \frac{ \text { 2x mol } }{ \text { 1.1 L } } ]^2}{ \frac{ \text { 0.0722 -x mol } }{ \text { 1.1 L } } \times \frac{ \text { 0.02299 -x mol } }{ \text { 1.1 L } } }$

$0.090 = \frac{ 4x^2}{ (0.0722 -x) \times ( 0.02299 -x )}$

${ (0.0722 -x) \times ( 0.02299 -x )} = \frac{ 4x^2} { 0.090 }$

${ (0.0722 -x) \times ( 0.02299 -x )} =44.44x^2$

This is quadratic equation with solution

$x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}$

$x = \frac{-0.095211 \pm \sqrt {(0.095211)^2-4(43.44)(-0.00166)}}{2(43.44)}$

$x = \frac{-0.095211 \pm0.0545}{86.89}$

or

The negative value is discarded since the number of moles cannot be negative.

The equilibrium concentrations are

$[Cl_2O] = \frac{ \text { 0.02299 -x mol } }{ \text { 1.1 L } }= \frac{ \text { 0.02299 -0.00518 mol } }{ \text { 1.1 L } }=0.016 \: M$

$[HOCl] = \frac{ \text { 2x mol } }{ \text { 1.1 L } } = \frac{ \text { 2 (0.00518) mol } }{ \text { 1.1 L } }=0.0094 \: M$

(b)

1.5 mol pure HOCl is placed in the flask

2x moles of HOCl will give x moles of H2O and x moles of Cl2O

The equilibrium concentrations are

$[H_2O] = \frac{ \text { x mol } }{ \text { 1.7 L } }$

$[Cl_2O] = \frac{ \text { x mol } }{ \text { 1.7 L } }$

$[HOCl] = \frac{ \text { 1.5-2x mol } }{ \text { 1.7 L } }$

$K = \frac{[HOCl]^2}{[H_2O][Cl_2O]}$

$0.090 = \frac{[ \frac{ \text { 1.5-2x mol } }{ \text { 1.7 L } } ]^2}{ \frac{ \text {x mol } }{ \text { 1.7 L } } \times \frac{ \text { x mol } }{ \text { 1.7 L } } }$

$0.090 = \frac{ (1.5-2x)^2}{ (x) \times ( x )}$

$0.090 = \frac{ (1.5-2x)}{ x}$

$x=0.718$

The equilibrium concentrations are

$[H_2O] = \frac{ \text { x mol } }{ \text { 1.7 L } }= \frac{ \text { 0.718 mol } }{ \text { 1.7 L } }=0.42 \: M$

$[Cl_2O] = \frac{ \text { x mol } }{ \text { 1.7 L } }= \frac{ \text { 0.718 mol } }{ \text { 1.7 L } }=0.42 \: M$

$[HOCl] = \frac{ \text { 1.5-2x mol } }{ \text { 1.7 L } }= \frac{ \text { 1.5-2(0.718) mol } }{ \text { 1.7 L } }=0.038 \: M$