7. The reaction: has Kp = 45.9 at 763 K. A particular equilibrium mixture at tha

Question

7. The reaction: has Kp = 45.9 at 763 K. A particular equilibrium mixture at that temperature contains gaseous Hl at a partial pressure of 4.00 atm and hydrogen gas at a partial pressure of 0.21 3 atm. What is the partial pressure of l2? 8. At a given temperature, the reaction above is at equilibrium when [CS2] = 0.050 M, [Cl2] = 0.25 M, [CCl4] = 0.15 M, and [S2Cl2] = 0.35 M. What would be the direction of the reaction when the reactants and products have the following concentrations: CS2 = 0.15 M, Cl2 = 0.18 M, CCI4 = 0.29 M, and S2Cl2 = 0.21 M?

Explanation / Answer

H2+I2---->2HI

The equilibrium constant Kp= [HIpp]2/ [H2pp *I2PP]

I2PP= [HIpp]2/(Kp*H2PP) (1)

HIpp= Partial pressure of HI =4 atm

H2PP= Partial pressure of Hydrogen= 0.213

I2PP= Partial pressure of Iodine= ?

Partial pressure of Iodine=

Substituting the values of Kp, H2PP and HIPP in Equation (1) gives the Partial pressure of Iodine I2PP.

Partial pressure of Iodine = (4)2/(0.213*45.9) =1.63 atma.

B is corect answer

Given that Kp=45.9

Partial pressue of HI = 4.59 atm

and Partial pressure of Hydrogen =0.213 atm

b) The reaction is CS2+3Cl2----->CCl4+S2Cl2

The moles of reactant at equlibrium = 0.05+0.25*3= 0.8M

Moles of products= 0.29+0.21 =0.5M

When the concentration were changed

CS2=0.15 M, Cl2=0.18M Total moles =0.15+3*0.18= 0.69

Total moles of product remained at 0.29+0.21 =0.5

There is a decrease in number of moles of reactant side when there is a change in concentration. As per Le-chatlier principle, the reaction shifts so as to compensate the effect of reducing moles. There is an increase in number of moles of reaction to the left.

Hence, the reaction shifts to the left.

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