Chemistry, 12.12.2019 01:31 parisw8

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 torr, 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.

Answers: 2

Show answers

Answers

Answer from: orlando19882000

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.

Answer from: Quest

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

Answer from: Quest

The answer would be true

Another question on Chemistry

Chemistry, 22.06.2019 03:10

The covalent compound acetylene, which is the fuel of the oxyacetylene torch used by welders, has the molecular formula c2h2. the covalent compound benzene, a commercial solvent, has the molecular formula c6h6 each of these covalent compounds contains carbon and hydrogen atoms in a one-to-one ratio. would it be correct to write the chemical formulas of each as ch? explain.

Answers: 1

Answer

Chemistry, 22.06.2019 12:30

How many grams of magnesium metal will react completely with 8.3 liters of 5.5m hcl? show all work

Answers: 1

Answer

Chemistry, 22.06.2019 20:30

Which states of matter have particles that move independently of one another with very little attraction?

Answers: 1

Answer

Chemistry, 22.06.2019 22:30

What is the value of the standard enthalpy of formation of an element in its most stable form?