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wolverine [178]
3 years ago
14

The temperature of 200 mL of gas

Chemistry
1 answer:
Elden [556K]3 years ago
6 0

Answer:

691.6 torr

Explanation:

Given data:

Initial temperature = 273 K

Initial pressure = 1 atm

Final temperature = -25 °C (-25 + 273 = 248 k)

Final pressure = ?

Solution:

P₁/T₁ = P₂/T₂

P₁ = Initial pressure

T₁ = Initial temperature

P₂ = Final pressure  

T₂ = Final temperature

Now we will put the values in formula.

P₁/T₁ = P₂/T₂

P₂ = P₁T₂/T₁  

P₂ = 1 atm × 248 K / 273 k

P₂ =248  atm. K / 273 K

P₂ = 0.91 atm  

In torr:

0.91  × 760 = 691.6 torr

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telo118 [61]

This is an incomplete question, here is a complete question.

Hydrogen and iodine react to form hydrogen iodide, like this:

H_2(g)+I_2(g)\rightarrow 2HI(g)

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

Compound            Pressure at equilibrium

H_2                                   61.8 atm

I_2                                    46.5 atm

HI                                  52.3 atm

Calculate the value of the equilibrium constant K_p for this reaction. Round your answer to 2 significant digits.

Answer : The value of equilibrium constant K_p for this reaction is, 0.952

Explanation :

The given chemical reaction :

H_2(g)+I_2(g)\rightarrow 2HI(g)

The expression of K_p for above reaction follows:

K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}

We are given:

P_{H_2}=61.8atm

P_{I_2}=46.5atm

P_{HI}=52.3atm

Putting values in above equation, we get:

K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952

Therefore, the value of equilibrium constant K_p for this reaction is, 0.952

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zloy xaker [14]

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