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Ivanshal [37]
3 years ago
12

An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To

rr, respectively. A quantity of Cl 2 ( g ) is injected into the mixture, and the total pressure jumps to 263.0 Torr. The appropriate chemical equation is PCl 3 ( g ) + Cl 2 ( g ) − ⇀ ↽ − PCl 5 ( g ) Calculate the new partial pressures after equilibrium is reestablished.
Chemistry
1 answer:
antoniya [11.8K]3 years ago
7 0

Answer: The new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}         ........(1)

We are given:

P_{PCl_5}=217.0torr

P_{PCl_3}=13.2torr

P_{Cl_2}=13.2torr

Putting values in above equation, we get:

K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24

Now we have to calculate the new partial pressure of Cl_2.

P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}

217.0torr+13.2torr+P_{Cl_2}=263.0torr

P_{Cl_2}=32.8torr

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of Cl_2.

Now, the equilibrium is shifting to the reactant side. The equation follows:

                       PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initial:             13.2         32.8            217.0

At eqm:         13.2-x      32.8-x         217.0+x

Putting values in expression 1, we get:

1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38

Neglecting the 40.4 value of 'x'  because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of PCl_5 = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of PCl_3 = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of Cl_2 = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

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Pete Gannett

 · 

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Ph.D. Chemistry, University of Wisconsin-Madison, (1982)2y

Seems to be an ideal gas law question. The relevant equation is:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles of gas, R is the gas constant (0.082 atm-L/mole-deg K), and T is temperature in Kelvins. STP means standard temperature and pressure and this is taken as 1 atm and 0º C or 273 K.

To calculate the number of molecules we will use the constant 6.023 * 10^23 molecules/mole and, therefore, we will need to know the number of moles (n). So, first we’ll rearrange the gas law equation, isolating ’n’ and then put the numbers in.

n = PV/RT = 1 * 1 / (0.082)(273) = 0.0447 moles

So, to calculate the number of molecules, multiple this by the number of molecules in a mole and you get:

# molecules of nitrogen in 1 Liter at STP = 6.023 * 10^23 molecules/mole * 0.0447 moles = 2.6905 * 10^22 molecules

Note, it does not matter what the gas is.

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