Question

The reaction of nitrogen gas with hydloric acid is as follows:

N_{2 (g)} + 6 HCl _(g)  

Answer 1

N2(g) + 6HCl(g)   <------> 2NH3(g) + 3Cl2(g)

(A) If volume is increased then equilibrium shifts to the left if delta(ng) is negative and to the right if delta(ng) is positive

delta(ng) = Moles of gaseous products - moles of gaseous reactants

delta(ng) = 5 - 7 = -2

So equilibrium proceeds to the left

(B)

If amount of Nitrogen is doubled then the equilibrium will shift to the right because more Nitrogen is available which will react with HCl to form more of product.

(C)

Since deltaH is positive so the reaction is endothermic i.e heat is absorbed in the system. So an increase of heat shifts the equilibrium towards right and more product is formed.

2.

I2(g)        +           Cl2(g)      <-------------------->        2ICl(g)

   4                            4                                                   0 (Initial concentration)

   -x                            -x                                                +2x (Change)

4 - x                        4 - x                                               2x ( Final concentration)

Now,

Kc = [ICl]^2 / [I2] * [Cl2]

0.10 = (2x)^2 / (4-x)*(4-x)

0.10 = (2x)^2 / (4-x)^2

0.10 = (2x / 4-x)^2

So,

(2x / 4-x) = square root (0.10)

( 2x / 4-x ) = 0.316              or      (2x / 4-x) = - 0.316

2x = 1.264 - 0.316x    or 2x = - 1.264 + 0.316x

2.316x = 1.264                 or         1.684x = - 1.264

x = 0.546 M                        or             x = - 0.75 ( Not possible because concentration cannot be negative)

So, x = 0.546 M

Final [ICl] = 2x = 2 * 0.546 M = 1.092 M