A catalytic converter in an automobile uses a palladium or platinum catalyst to

Question

A catalytic converter in an automobile uses a palladium or platinum catalyst to convert carbon monoxide gas to carbon dioxide according to the following reaction:
2CO(g)+O2(g)2CO2(g)
A chemist researching the effectiveness of a new catalyst combines a 2.0 : 1.0 mole ratio mixture of carbon monoxide and oxygen gas (respectively) over the catalyst in a 2.45L flask at a total pressure of 755 torr and a temperature of 552 C. When the reaction is complete, the pressure in the flask has dropped to 556 torr .

What percentage of the carbon monoxide was converted to carbon dioxide?

Explanation / Answer

Reaction :

2CO(g)+O2(g) -- > 2CO2(g)

Mol ration of CO2 : O2 is 2.0 : 1.0

Volume of tank = 2.45 L

Total pressure =755 torr

T =552 Torr

Final pressure at equilibrium = 556 torr

Solution :

We have to find percent of CO converted to CO2

Lets assume moles of CO taken =2 mol and that of O2 is 1

Since mole ratio is also 2 : 1 so yield of both is important in product formation

Lets find out how much CO2 is formed.

For this we use following equation

Here V and T is constant

So we can use following equation.

P1 / n1 = p2 / n2

Let p1 = 755 torr

P2= 556 torr

n 1 = total moles taken = 2 + 1 = 3 mol

Lets plug in the values

n 2 = 556 x 3 / 755 = 2.2093 mol

Now there are 2.2093 moles are present in this tank

Lets assume fraction of CO reacted to be y

So the moles of CO2 formed

= Number of moles of CO x y

So total moles of CO2 in the tank at equilibrium = 2.0 y

Now total moles of O2 = 2.0 y x 1 mol O2 / 2 mol CO2

= 1.0 y

And moles of CO at equilibrium = 2.0 – 2.0 y

Total moles present at equilibrium = 2.0 y + (1.0 y) + 2.0y

Total moles we got =2.2093

Lets find y

2.0 y + (1.0 y) + 2.0y = 2.2093

5.0y = 2.2093

y = 0.4419

Percent conversion of CO to CO2 is

= 0.442 x 100

= 44.2 %

So the percent conversion of CO is 44.2 %

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