Chemistry, asked by jagadeesh005, 9 months ago

A vessel at 1000K contains CO2 with a pressure of 0.5 atm. Some of the co2 is converted into CO on the addition of graphite. if the total pressure at equilibrium is 0.8 atm, The value of Kp is

Answers

Answered by abdul9838

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$\bf \red{hey \: user \: here \: is \: answer} \\ \\ \bf \red{ \huge \: solution} \\ \\$

According to the question

The reaction is happening at constant temperature and at constant volume

Now

First writing the balance chemical equation

CO2(g)+C(s)=====>2Co(g)

Initially partial pressure of cO2=0. 5atm

====> Pco2=0. 5 atm

And

Initially partial pressure of Co =0 atm

According to the question

Some of CO2 is converted into Co

Therefore let converted CO2 be x

Now

At equilibrium

Partial pressure of CO2=(0. 5-x) atm

Partial pressure of Co=2x atm

Total pressure is given at 0. 8 atm

Pt=0. 8 atm

Now

Pt=Pco2+Pco

0. 8=(0. 5-x)+2x

$\bf \red{0.8 = 0.5 - x + 2x} \\ \\ \bf \red{0.8 - 0.5 = x} \\ \\ \bf \red{x = 0.3 \: atm}$

Now

Partial pressure of CO2=(0. 5-x)=(0. 5-0. 3)=0. 2

==> Pco2=0. 2 atm

Partial pressure of Co=2x=2×0. 3=0. 6 atm

Pco=0. 6 atm

Now

Before writing the expression for equilibrium constant(Kp )we have to keep in mind that solids are not added in expression.

$\bf \red{Kp = \frac{[CO]^{2} }{[CO2]} } \\ \\ \bf \red{Kp= \frac{ {(0.6)}^{2} }{0.2} } \\ \\ \bf \red{Kp = \frac{0.36}{0.2} } \\ \\ \bf \red{Kp= \frac{0.36 \times 100}{0.2 \times 100}} \\ \\ \bf \red{Kp= \frac{ \cancel{36}}{ \cancel{20}} } \\ \\ \bf \red{Kp= \frac{9}{5} } \\ \\ \bf \red{Kp= 1.8 \: atm}$

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A vessel at 1000K contains CO2 with a pressure of 0.5 atm. Some of the co2 is converted into CO on the addition of graphite. if the total pressure at equilibrium is 0.8 atm, The value of Kp is​

Answers

According to the question

The reaction is happening at constant temperature and at constant volume

Now

First writing the balance chemical equation

CO2(g)+C(s)=====>2Co(g)

Initially partial pressure of cO2=0. 5atm

====> Pco2=0. 5 atm

And

Initially partial pressure of Co =0 atm

According to the question

Some of CO2 is converted into Co

Therefore let converted CO2 be x

Now

At equilibrium

Partial pressure of CO2=(0. 5-x) atm

Partial pressure of Co=2x atm

Total pressure is given at 0. 8 atm

Pt=0. 8 atm

Now

Pt=Pco2+Pco

0. 8=(0. 5-x)+2x

Now

Partial pressure of CO2=(0. 5-x)=(0. 5-0. 3)=0. 2

==> Pco2=0. 2 atm

Partial pressure of Co=2x=2×0. 3=0. 6 atm

Pco=0. 6 atm

Now

Before writing the expression for equilibrium constant(Kp )we have to keep in mind that solids are not added in expression.

A vessel at 1000K contains CO2 with a pressure of 0.5 atm. Some of the co2 is converted into CO on the addition of graphite. if the total pressure at equilibrium is 0.8 atm, The value of Kp is