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tia_tia [17]
3 years ago
5

A mixture of CO (g) and excess O2(g) is placed in a 1.0 L reaction vessel at 100.0C and a total pressure of 1.50 atm. The CO is

ignited and it burns according to the balanced equation below. After reaction, the vessel is cooled to 100.0C and the pressure of the mixture of product gases (O2(g) , CO2(g)) is 1.40 atm. What is the partial pressure of of CO2 in the product mixture
Chemistry
1 answer:
umka2103 [35]3 years ago
7 0

Answer:

Partial pressure of of CO₂ in the product mixture is 0,20atm

Explanation:

The balance equation is:

2CO(g) + O₂(g) → 2CO₂(g)

Total pressure of CO(g) and O₂(g) gases before reaction at 100,0°C and 1,0L is 1,50 atm. You can say:

X₁ + Y₁  = 1,50atm <em>(1)</em>

Where X₁ is initial partial pressure CO and Y₁ is initial partial pressure of O₂

After reaction partial pressures are:

X₂ = X₁ - 2n = 0; <em>2n = X₁</em>

Y₂ = Y₁ - n

Z₂ = 2n

Where Z₂ is final partial pressure of CO₂

After reaction pressure at 100,0°C and 1,0L is 1,40 atm, that means:

1,40 atm = (Y₂ + Z₂)

1,40 atm = Y₁ - n + 2n

1,40atm = Y₁ + n

1,40 atm = Y₁ + X₁/2 <em>(2)</em>

Replacing (1) in (2)

1,40 atm = 1,50atm - X₁ + X₁/2

-0,10 atm = - X₁/2

<em>0,20 atm = X₁</em>.

As 2n = X₁; 2n =<em> Z₂ = 0,20 atm</em>

I hope it helps!

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Now put all the given values in Q_1, we get:

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Q_2 = amount of heat absorbed = ?

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Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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