Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Chemical Principles - Solved Quiz, Exams of Chemical Principles

Solved quiz for Chemical Principles course in class of Dr. Brian Lamp at Truman State University

Typology: Exams

2018/2019

Uploaded on 04/15/2019

atifarifasif
atifarifasif 🇵🇰

4.2

(14)

33 documents

1 / 2

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
CHEM 130 Name _____________________________________________
Quiz 9 – November 17, 2017
Complete the following problems. Write your final answers in the blanks provided.
1. At 25
o
C, 0.760 mol SO
3
is placed in an otherwise empty 5.00 L container. When equilibrium is
reached, 0.160 mol of O
2
is present. What is K
c
at this temperature? (8 pts)
2SO
3
(g) 2SO
2
(g) + O
2
(g)
There are several ways to approach this. In the end, we need equilibrium concentrations of all
species.
2SO
3
2SO
2
+
O
2
I 0.760 mol/5 L = 0.152 M 0 0
C
-2x +2x +x
E
0.152-2(0.032) =0.088 M
0+2(0.032) =0.064 M
0.160 mol/5L =
0.
0
320 M
K
c
=
[SO
2
]
2
[O
2
]
=
(0.0640)
2
(0.0320)
=
0.0169
[SO
3
]
2
(0.088)
2
Answer_______0.0169________________
2. Determine K
c
for the reaction: 2NOCl(g) N
2
(g) + O
2
(g) + Cl
2
(g) from the following data at 298K:
(8 points) ½ N
2
(g) + O
2
(g) NO
2
(g) K
c
= 1.0 x 10
-
9
NOCl(g) + ½ O
2
(g) NO
2
Cl(g) K
c
= 1.1 x 10
2
NO
2
(g) + ½ Cl
2
(g) NO
2
Cl(g) K
c
= 0.3
We need to rearrange reactions to make our target reaction:
2(NO
2
(g) ½ N
2
(g) + O
2
(g)) K
c
= 1/(1.0 x 10
-
9
)
2
2(NOCl(g) + ½ O
2
(g) NO
2
Cl(g)) K
c
= (1.1 x 10
2
)
2
2(NO
2
Cl(g) NO
2
(g) + ½ Cl
2
(g)) K
c
= 1/(0.3)
2
So, the sum of the reactions is:
2NO
2
+ 2NOCl + O
2
+ 2NO
2
Cl N
2
+ 2O
2
+ 2NO
2
Cl + 2NO
2
+ Cl
2
2NO
2
Cl N
2
+ O
2
+ Cl
2
K
c
=
(1.1 x 10
2
)
2
=
1.3 x10
23
(1.0 x 10
-
9
)
2
(0.3)
2
Answer______1.3 x10
23
_________________
pf2

Partial preview of the text

Download Chemical Principles - Solved Quiz and more Exams Chemical Principles in PDF only on Docsity!

CHEM 130 Name _____________________________________________ Quiz 9 – November 17, 2017

Complete the following problems. Write your final answers in the blanks provided.

  1. At 25oC, 0.760 mol SO 3 is placed in an otherwise empty 5.00 L container. When equilibrium is reached, 0.160 mol of O 2 is present. What is Kc at this temperature? (8 pts) 2SO 3 (g) 2SO 2 (g) + O 2 (g)

There are several ways to approach this. In the end, we need equilibrium concentrations of all species.

2SO 3 2SO 2 + O 2 I 0.760 mol/5 L = 0.152 M 0 0 C -2x +2x +x E 0.152-2(0.032) =0.088 M 0+2(0.032) =0.064 M 0.160 mol/5L = 0. 0 320 M =x

Kc = [SO 2 ]^2 [O 2 ] = (0.0640)^2 (0.0320) = 0. [SO 3 ]^2 (0.088)^2

Answer_______ 0.0169 ________________

  1. Determine Kc for the reaction: 2NOCl(g) N 2 (g) + O 2 (g) + Cl 2 (g) from the following data at 298K: (8 points) ½ N 2 (g) + O 2 (g) NO 2 (g) Kc = 1.0 x 10-^9 NOCl(g) + ½ O 2 (g) NO 2 Cl(g) Kc = 1.1 x 10^2 NO 2 (g) + ½ Cl 2 (g) NO 2 Cl(g) Kc = 0.

We need to rearrange reactions to make our target reaction:

2(NO 2 (g) ½ N 2 (g) + O 2 (g)) Kc = 1/(1.0 x 10-^9 )^2 2(NOCl(g) + ½ O 2 (g) NO 2 Cl(g)) Kc = (1.1 x 10^2 )^2 2(NO 2 Cl(g) NO 2 (g) + ½ Cl 2 (g)) Kc = 1/(0.3)^2

So, the sum of the reactions is:

2NO 2 + 2NOCl + O 2 + 2NO 2 Cl N 2 + 2O 2 + 2NO 2 Cl + 2NO 2 + Cl 2 2NO 2 Cl N 2 + O 2 + Cl 2

Kc = (1.1 x 10^2 )^2 = 1.3 x10^23 (1.0 x 10-^9 )^2 (0.3)^2

Answer______ 1.3 x10^23 _________________

  1. Phosphorus pentachloride decomposes according to the chemical equation below. A 0.280 mol sample of PCl 5 (g) is injected into an empty 4.00 L reaction vessel held at 250°C. Calculate the concentrations of PCl 5 (g) and PCl 3 (g) at equilibrium. PCl 5 (g) PCl 3 (g) + Cl 2 (g) Kc = 1.80 at 250°C

PCl 5 PCl 3 + Cl 2 I 0.280 mol/4 L = 0.0700 M 0 0 C -x +x +x E 0.0700-x x x

Kc = [PCl 3 ][Cl 2 ] = (x)(x) [PCl 5 ] 0.0700-x

Kc = x^2 0.0700-x

(0.0700-x)Kc = x^2 0.0700Kc-xKc = x^2 x^2 + Kcx – 0.0700Kc = 0 Inserting 1.80 for Kc x^2 + 1.80x – 0.126 = 0 Using the quadratic equation with a = 1, b = 1.80 and c = -0.126 gives: x = 0.067 or -1. Since we’ve defined x as a concentration, a negative value makes no chemical sense, therefore the appropriate value for x is 0.067 or [PCl 3 ] = [Cl 2 ] = 0.0675 M and [PCl 5 ] = 0.0700-0.0675 = 0.0025 M

Answer_ [PCl 3 ] 0.0675 M and [PCl 5 ] = 0.0025 M_

Possibly Useful Information

2 1

2 1 x x

y y x

y slope m −

2 a

b b 4 ac x

− ±^2 −

= R = 0.08206 L atm mol

-1 K-

R = 8.314 J mol-1^ K-

pV = nRT ∆G = -RTlnK Kp = Kc(RT)∆n