Chemistry, asked by sonalgpatel217, 1 year ago

ch3coch3 = ch3-ch3 + co ,initial pressure of ch3coch3 is 100 mm. at eqlb mole fraction of co is 1/4 . find value of kp​

Answers

Answered by Anonymous
94

Answer -

\sf{K_p\:=\: 50}

\rule{200}2

Explanation -

Given:-

  • \sf{CH_3COCH_3\:=\:CH_3CH_3\:+\:CO}
  • Initial pressure of \sf{CH_3COCH_3} = 100 mm
  • Mole fraction of \sf{CH_3CH_3\:=\:CO\:=\:\frac{1}{4}}

Find:-

Value of \sf{K_p}

Solution:-

\sf{CH_3COCH_3\:=\:CH_3CH_3\:+\:CO}

and \sf{CH_3CH_3\:=\:CO\:=\:\dfrac{1}{4}}

\implies\:\sf{CH_3COCH_3\:\longleftrightarrow\:CH_3CH_3\:+\:CO}

It is given that, initially\sf{CH_3COCH_3} is 100 mm.

\implies\:\sf{CH_3COCH_3\:\longleftrightarrow\:CH_3CH_3\:+\:CO}

At Initial) 100mm............... 0 . .............. 0

At Equilibrium) (100 - C).... C ................ C

When non-volatile solute is added to a solvent, then vapour pressure of solvent decreases because some of the space is occupied by solute particles which create hindrance to the formation of particles.

Vapour pressure of solution is equal to the partial pressure of solvent.

So,

\Rightarrow\:\sf{x_A\:=\:\dfrac{P_A}{P_T}}

Here -

  • \sf{x_A} is mole fraction
  • \sf{P_A\:=\:C} is partial pressure
  • \sf{P_T} is total pressure

\implies\:\sf{C\:=\:P_T\:\times\:\dfrac{1}{4}}

\implies\:\sf{C\:=\:\dfrac{P_T}{4}} ...(1)

Similarly,

\implies\:\sf{100\:-\:C\:=\:\dfrac{P_T}{4}} ...(2)

Add equation (1) & (2)

\implies\:\sf{C\:+\:100\:-\:C\:=\:\dfrac{P_T}{4}\:+\:\dfrac{P_T}{4}}

\implies\:\sf{100\:=\:\dfrac{2P_T}{4}}

\implies\:\sf{P_T\:=\:200\:mm}

Now,

\implies\:\sf{C\:=\:\dfrac{P_T}{4}}

\implies\:\sf{C\:=\:\dfrac{200}{4}}

\implies\:\sf{C\:=\:50}

\sf{K_p\:=\:\dfrac{C^2}{100\:-\:C}}

\implies\:\sf{K_p\:=\:\dfrac{(50)^2}{100\:-\:50}}

\implies\:\sf{K_p\:=\:\dfrac{(50)^2}{50}}

\implies\:\sf{K_p\:=\:50}

Answered by Anonymous
46

Answer:

  \large\bold\red{{K}_{p} = 50}

Explanation:

Given,

A chemical Equation,

\bold{CH_{3}COCH_{3} \leftrightarrow CH_{3}- CH_{3}  + CO}

Initial Pressure of CH_{3}COCH_{3} = 100 mm

At equlibrium,

Mole fraction of CO = \frac{1}{4}

Since,

Both CO and CH_{3}-CH_{3} have equal moles.

Therefore,

Mole fraction of CH_{3}-CH_{3} = \frac{1}{4}

Now,

Let the amount of CH_{3}COCH_{3} dissociated in terms of partial pressure is ‘c’ mm.

Therefore,

We have the following conditions.

\bold\red{Note:-} Refer to the attachment for Equation.

Now,

We have,

Partial Pressure of,

  • CH_{3}COCH_{3} = (100-c)\:mm
  • CH_{3}-CH_{3}= c\:mm
  • CO = c\:mm

Let the Total Pressure = {P}_{total}

We know that,

\large\boxed{\bold{{P}_{partial} = {P}_{total} \times (Mole\: Fraction)}}

Therefore,

We have,

  =  > c = {P}_{total} \times  \frac{1}{4}  \\  \\  =  > c =  \frac{{P}_{total}}{4}  \:  \:  \:  \:  \:  \:  \:  \:  ...........(1)

Also,

 =  > 100 - c = {P}_{total} \times  \frac{1}{4}  \\  \\  =  > 100 - c =  \frac{{P}_{total}}{4}  \:  \:  \:  \:  \:  \:  \: ...........(2)

Adding the Equation (1) and (2),

We get,

 =  > 2 \times  \frac{{P}_{total}}{4}  = 100 \\  \\  =  > {P}_{total} = 200 \: mm

Therefore,

We have,

 =  > c =  \frac{200}{4}   \\  \\  =  > c = 50 \: mm

Now,

We know that,

 \large \boxed{ \bold{{K}_{p} =  \frac{ {c}^{2} }{(100 - c)} }}

Therefore,

Putting the values,

We get,

 =  > {K}_{p} =  \frac{50 \times 50}{(100 - 50)}  \\  \\  =  > {K}_{p} =  \frac{50 \times  \cancel{50}}{ \cancel{50}}  \\  \\  =  >  \large \bold{{K}_{p} = 50}

Attachments:

Anonymous: Write units too
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