Three gases (NO, Cl_2, and NOCI) arc placed into a 1.0L reaction vessel at 25 de

ID: 931643 • Letter: T

Question

Three gases (NO, Cl_2, and NOCI) arc placed into a 1.0L reaction vessel at 25 degreeC.The initial partial pressures of the gases are 1.5 atm for NO, 2.2 atm for Cl_2, and 3.3 atm for NOCl. Which of the following would be true for these conditions based on the equilibrium reaction below at 25 degreeC? 2 NOCI (g) 2 NO (g) + Cl_2 (g) K_p = 0.863 The system is at equilibrium. NO will be consumed in order to reach equilibrium The partial pressure of NOCI will increase as equilibrium is reached The partial pressure of Cl; will increase as equilibrium is reached. None of the above are true.

Explanation / Answer

At equilibrium,

Kp=P(NO)^2*P(Cl2)/P(NOCl)^2 where P represents partial pressure of gases

plug in the values of the partial pressures,

Q=(1.5 atm)^2 *(2.2 atm)/(3.3 atm)=1.5

As the reaction quotient Q is not equal to Kp so the reaction has not reached the equilibrium .

To calculate equilibrium partial pressures of all the gases, prepare ICE table

pNOcl

pNO

pCl2

Initial

3.3atm

1.5atm

2.2

change

-x

equilibrium

3.3-x

1.5+x

2.2+x

Kp=P(NO)^2*P(Cl2)/P(NOCl)^2 =(1.5-x)^2 *(2.2-x)/(3.3-x)^2

0.863=(1.5+x)^2 *(2.2+x)/(3.3-x)^2

0.863=(2.25+x^2+3x ) (2.2+x)/(10.89+x^2-6.6x)

0.863=4.95+2.25x +2.2x^2+x^3+6.6x+3X^2/(10.89+x^2-6.6x)

0.863=4.95+8.85x+5.2X^2+X^3/(10.89+x^2-6.6x)

Ignore all higher x terms as x is small ,

0.863=(4.95+8.85x)/ (10.89-6.6x)

9.398-5.69x=4.95+8.85x

-14.54x=-4.448

X=0.306 atm

Equilibrium partial pressures,

pNO =1.5+x=1.5+0.306=1.806 atm

pcl2=2.2+x=2.2+0.306=2.506 atm

pNOCl=3.3-x=3.3-0.306=2.99 atm

Answer=only D is correct