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Olin [163]
3 years ago
14

A gas that exerts a pressure of 537 torr in a container with a volume of 5.30 L will exert a pressure of 255 torr when transferr

ed to a container with a volume of ? L
Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
4 0

Answer:- The gas needs to be transferred to a container with a volume of 11.2 L.

Solution:- From Boyle's law. "At constant temperature, Volume is inversely proportional to the pressure."

It means, the volume is decreased if the pressure is increased and vice versa.

Here, the Pressure is decreasing from 537 torr to 255 torr. So, the volume must increase and calculated by using the equation:

P_1V_1=P_2V_2

Where, P_1 is initial pressure and P_2 is final pressure. Similarly, V_1 is initial volume and V_2 is final volume.

Let's plug in the values in the equation:

(537 torr)(5.30 L) = (255 torr)(V_2)

V_2=(\frac{537 torr*5.30 L}{255 torr})

V_2 = 11.2 L

So, the new volume of the container needs to be 11.2 L.

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The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate
Karolina [17]

<u>Answer:</u> The value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

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Relation of K_p with K_c is given by the formula:

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

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K_p = equilibrium constant in terms of partial pressure = 3.45

K_c = equilibrium constant in terms of concentration = ?

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

T = temperature = 500 K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-1=1

Putting values in above equation, we get:

3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084

The equation used to calculate concentration of a solution is:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}

Initial moles of (CH_3)_3CCl(g) = 1.00 mol

Volume of the flask = 5.00 L

So, \text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M

For the given chemical reaction:

                (CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)

Initial:               0.2                    -                        -

At Eqllm:          0.2 - x               x                       x

The expression of K_c for above reaction follows:

K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}

Putting values in above equation, we get:

0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

[(CH_3)_2C=CH]=x=0.094M

[HCl]=x=0.094M

[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M

Hence, the value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

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