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mamaluj [8]
3 years ago
11

A student is working with a 4.95 L sample of carbon dioxide at 22.8°C and 0.956 atm. The student compresses the gas to 1.75 L an

d increases the temperature of the gas to 65.3°C. What will the resulting pressure of the CO 2 gas be after the student made these changes?
Chemistry
1 answer:
myrzilka [38]3 years ago
5 0

 The resulting  pressure   is 3.093 atm


   <u><em>calculation</em></u>

The  resulting pressure is calculated  using the combined gas equation

That is   P₁V₁/T₁ =P₂V₂/T₂   where,

P₁=0.956 atm

V₁=4.95 L

T₁=22.8 c  in to kelvin= 22.8 +273 =295.8 k

P₂=? atm

V₂=1.75 L

T₂=65.3 c  into kelvin = 65.3 +273  =338.3 K

make P₂ the subject  of the formula by multiplying  both side of the formula  by T₂/V₂

P₂ =T₂P₁V₁ / V₂T₁

P₂  = [(338.3 k x 0.956 atm  x 4.95 L) / ( 1.75 L x 295.8 K)] =3.093 atm


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The equilibrium constant Kc for the decomposition of phosgene, COCl2, is 4.63x10^-3 at 527 C. Calculate the equilibrium partial
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<u>Answer:</u> The partial pressure of the CO,Cl_2\text{ and }COCl_2 are 0.352 atm, 0.352 atm and 0.408 atm respectively.

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K_c=4.63\times 10^{-3}

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Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

Where,

K_p = equilibrium constant in terms of partial pressure = ?

K_c = equilibrium constant in terms of concentration = 4.63\times 10^{-3}

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 527^oC=527+273=800K

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Putting values in above equation, we get:

K_p=4.63\times 10^{-3}\times (0.0821\times 800)^{1}\\\\K_p=0.304

The chemical reaction for the decomposition of phosgene follows the equation:

                   COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

At t = 0          0.760            0     0  

At t=t_{eq}      0.760-x          x      x

The expression for K_p for the given reaction follows:

K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}

We are given:

K_p=0.304\\p_{COCl_2}=0.760-x\\p_{CO}=x\\p_{Cl_2}=x

Putting values in above equation, we get:

0.304=\frac{x\times x}{0.760-x}\\\\x=0.352,-0.656

Negative value of 'x' is neglected because partial pressure cannot be negative.

So, the partial pressure for the components at equilibrium are:

p_{COCl_2}=0.760-0.352=0.408atm\\\\p_{CO}=0.352atm\\\\p_{Cl_2}=0.352atm

Hence, the partial pressure of the CO,Cl_2\text{ and }COCl_2 are 0.352 atm, 0.352 atm and 0.408 atm respectively.

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