Question

An equilibrium mixture consists of 0.31 atm NO2(g) and 0.45 atm N2O4(g). 2 NO2(g) <—> N2O4(g) A chemist adds enough NO2 to this mixture to increase the equilibrium partial pressure of NO2(g) to 0.40 atm. What additional pressure of NO2 did she add to the container? Assume that the temperature and volume remain constant.

Answer #1

** 2 NO2(g) <—>
N2O4(g)**

**equi 0.31 atm 0.45
atm**

**add
x
-**

**change
-2y
+y**

**final 0.4 atm
0.45+y**

**Kp = pN2O4/pNO2^2**

** = 0.45/(0.31)^2**

** = 4.68**

**final**

**4.68 = (0.45+y)/(0.4^2)**

**y = 0.3**

**so that, 0.31+x-2y = 0.4**

**
0.31+x-2*0.3 = 0.4**

** x = 0.69 atm**

**additional pressure of NO2 did she add to the container=
x = 0.69 atm**

An equilibrium mixture contains N2O4, (P= 0.27atm ) and
NO2 (P= 1.1 atm ) at 350 K. The volume of the container is
doubled at constant temperature.
A) Calculate the equilibrium pressure of
N2O4 when the system reaches a new equilibrium.
B) Calculate the equilibrium pressure of
NO2 when the system reaches a new equilibrium.

An equilibrium mixture contains N2O4,
(P= 0.28 atm ) and NO2 (P= 1.0 atm )
at 350 K. The volume of the container is doubled at constant
temperature.
Part A
Calculate the equilibrium pressure of N2O4
when the system reaches a new equilibrium.
Express your answer using two significant figures.
Part B
Calculate the equilibrium pressure of NO2 when the
system reaches a new equilibrium.
Express your answer using two significant figures.

A flask is charged with 1.500 atm
N2O4(g) and 1.00 atm
NO2(g) at 25 °C and the following equilibrium
is achieved:
N2O4(g)<------->
2NO2(g)
After equilibrium is reached, the partial pressure of NO2 is
0.512 atm. What is the equilibrium partial pressure of N2O4?

A flask is charged with 1.800 atm of N2O4(g)and 1.00
atm NO2(g) at 25 ∘C, and the following equilibrium is
achieved:
N2O4(g)⇌2NO2
After equilibrium is reached, the partial pressure of NO2 is 0.515
atm .
What is the equilibrium partial pressure of
N2O4?
Calculate the value of Kp for the
reaction
Calculate Kc for the
reaction.

A flask is charged with 1.800 atm of
N2O4(g) and 1.00 atm of
NO2(g) at 25 ∘C , and the following
equilibrium is achieved:
N2O4(g)⇌2NO2(g)
After equilibrium is reached, the partial pressure of
NO2 is 0.519 atm .
Part A:
What is the partial pressure of N2O4 at
equilibrium?
Part B:
Calculate the value of Kp for the
reaction.
Part C:
Calculate the value of Kc for the
reaction.

At 25°C, the equilibrium partial pressures of NO2 and N2O4 are
0.150 atm and 0.200 atm, respectively. If the volume is increased
by 1.60 fold at constant temperature, calculate the partial
pressures of the gases when a new equilibrium is established.
PNO2 = atm
PN2O4 = atm

A flask is charged with 1.800 atm of N2O4(g)and 1.00
atm NO2(g) at 25 ∘C, and the following equilibrium is
achieved:
N2O4(g)⇌2NO2
After equilibrium is reached, the partial pressure of NO2 is 0.517
atm .
1)What is the equilibrium partial pressure of N2O4?
Express your answer with the appropriate units.
2)Calculate the value of Kp for the
reaction.
3)Calculate Kc for the reaction.

A flask is charged with 1.550 atm of N2O4(g)and 1.00
atm NO2(g) at 25 ∘C, and the following equilibrium is
achieved:
N2O4(g)⇌2NO2
After equilibrium is reached, the partial pressure of NO2 is 0.519
atm .
1.What is the equilibrium partial pressure of N2O4?Express your
answer with the appropriate units.
2. Calculate the value of Kp for the
reaction.
3.Calculate the value of Kc for the reaction.

A flask is charged with 1.500 atm of N2O4(g) and 1.00 atm NO2(g)
at 25 ∘C, and the following equilibrium is achieved: N2O4(g)⇌2NO2
After equilibrium is reached, the partial pressure of NO2 is 0.514
atm .
Calculate Kc for the reaction.

At a particular temperature, Kp = 0.26 for the reaction below.
N2O4(g) equilibrium reaction arrow 2 NO2(g) (a) A flask containing
only N2O4 at an initial pressure of 4.9 atm is allowed to reach
equilibrium. Calculate the equilibrium partial pressures of the
gases. b) he volume of the container in part (a) is decreased to
one-half the original volume. Calculate the new equilibrium partial
pressures.

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