Consider the following reaction: 2SO2(g)+O2(g)⇌2SO3(g) Kp=0.355 at 950 K A 2.75−L reaction vessel at 950K initially contains 0.100 mol of SO2 and 0.100 mol of O2.Calculate the total pressure (in atmospheres) in the reaction vessel when equilibrium is reached.

Question

25-03-2020
Chemistry

contestada

Consider the following reaction: 2SO2(g)+O2(g)⇌2SO3(g) Kp=0.355 at 950 K A 2.75−L reaction vessel at 950K initially contains 0.100 mol of SO2 and 0.100 mol of O2.Calculate the total pressure (in atmospheres) in the reaction vessel when equilibrium is reached.

Respuesta :

Otras preguntas

A triathlete swims a distance of 3 kilometers in 1 hour, then bikes a distance of 50 kilometers in 3 hours, and finally then runs a distance of 12 kilometers in

help pls ( wrong answer report

In The Secret Garden, what does Mary do when she hears crying down the hall? She concludes that it must be a ghost since the estate is enchanted. She enlists Ma

You are paid $1380 for working 80 hours in two weeks at your full-time job. At this rate of pay, how much would you be paid if ypu worked 100 hours in a month

A 8-kg object is moving with a speed of 5 m/s at a height of 4 m. The kinetic energy of the object is approximately ____ J. *

NEED ANSWER NOWWW!!!look at the picture i below. How has the artist emphasized the figure of Rama? What elements and principles of art help capture the action o

water flowing from toilet to tap may be hard to swallow evaluation paragraph

Read this sentence. It is important for citizens to vote because it is his duty to choose the country's leaders. Which option corrects the pronoun mistake in th

Under what circumstances, if any, do you think a citizen in a representative democracy has the right to engage in civil disobedience? Explain your position.

The amount of a person’s paycheck is directly proportional to the number of hours worked. For working 11 hours, the pay is $79.75. Find the pay for 35 hours of

Alleei Alleei · Answer 1 · 2020-03-25T09:07:00+01:00

Answer : The total pressure (in atmospheres) in the reaction vessel when equilibrium reached is, 5.02 atm

Explanation :

First we have to calculate the initial pressure of [tex]SO_2[/tex] and [tex]O_2[/tex]

[tex]P_{SO_2}=\frac{n_{SO_2}RT}{V}[/tex]

[tex]P_{SO_2}=\frac{(0.100mol)\times (0.0821L.atm/mol.K)\times (950K)}{2.75L}[/tex]

[tex]P_{SO_2}=2.84atm[/tex]

and,

[tex]P_{O_2}=\frac{n_{O_2}RT}{V}[/tex]

[tex]P_{O_2}=\frac{(0.100mol)\times (0.0821L.atm/mol.K)\times (950K)}{2.75L}[/tex]

[tex]P_{O_2}=2.84atm[/tex]

The balanced chemical reaction is:

[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

Initial pressure 2.84 2.84 0

At eqm. (2.84-2x) (2.84-x) 2x

The expression of [tex]K_p[/tex] for above reaction follows:

[tex]K_p=\frac{(P_{SO_3})^2}{(P_{SO_2})^2\times P_{O_2}}[/tex]

Putting values in expression 1, we get:

[tex]0.355=\frac{(2x)^2}{(2.84-2x)^2\times (2.84-x)}[/tex]

x = 0.664

Now we have to calculate the total pressure.

Partial pressure of [tex]SO_3[/tex] = (2x) = 2(0.664) = 1.33 atm

Partial pressure of [tex]SO_2[/tex] = (2.84-2x) = [2.84-2(0.664)] = 1.51 atm

Partial pressure of [tex]O_2[/tex] = (2.84-x) = [2.84-0.664] = 2.18 atm

Total pressure = Partial pressure of [tex]SO_3[/tex] + Partial pressure of [tex]SO_2[/tex] + Partial pressure of [tex]O_2[/tex]

Total pressure = 1.33 + 1.51 + 2.18

Total pressure = 5.02 atm

Thus, the total pressure (in atmospheres) in the reaction vessel when equilibrium reached is, 5.02 atm

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-18T16:30:25+01:00

The total pressure is mathematically given as

T=5.02

What is the total pressure in the reaction vessel?

Generally, the equation for the pressure is mathematically given as

For SO2

[tex]P_{SO_2}=\frac{n_{SO_2}RT}{V}[/tex]

Therefore

[tex]P_{SO_2}=\frac{(0.100)* (0.0821)* (950K)}{2.75L}[/tex]

[tex]P_{SO_2}=2.84atm[/tex]

For O2

[tex]P_{O_2}=\frac{(0.100)* (0.0821)* (950K)}{2.75L}\\\\P_{O_2}=2.84atm[/tex]

Hence, The balanced equation is

2SO_2(g)+O_2(g) ---><---- 2SO_3(g)

Therefore using Kp equation

[tex]0.355=\frac{(2x)^2}{(2.84-2x)^2* (2.84-x)}[/tex]

x=0.67

Now solving partial pressure for SO3,SO2,O2 We have

PSO3= (2x) = 2(0.664)

PSO3= 1.33 atm

PSO2 = (2.84-2x)

PSO2 = 1.51 atm

PO= [2.84-0.664]

PO= 2.18 atm

Tottal pressure

T=1.33 + 1.51 + 2.18

T=5.02