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Gemiola [76]
3 years ago
12

To what volume will a 2.33 l sample of gas expand if it is heated from 30.0 ∘c to 300.0 ∘c?

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0
According to Charle's Law,
V/T =  constant
where 
V = volume
T =  absolute temperature.

The law holds when the pressure of an ideal gas is held constant while temperature changes.

V₁ = 2.33 L, initial volume
T₁ = 30°C = 30+273 K = 303 K, initial temperature
T₂ = 300°C = 300+273 K = 573 K, final temperature

The final volume, V₂, is given by
V₂/T₂ = V₁/T₁
V₂ = (573 K)*(2.33/303  L/K)
     =  4.406 L

Answer: 4.41 L

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Answer : The value of K_c for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

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As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

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The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

Now we have to calculate the value of K_c for the given reaction.

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

K_c=\frac{(0.6)^2}{(1)\times (1)}

K_c=0.36

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