Consider the reaction n2(g) + o2(g) â 2 no(g) k = 0.0025 a rigid container initially contains 8.00 atm of nitrogen and 5.00 atm of oxygen. what will be the partial pressure of nitrogen at equilibrium?

7.8458 atm

Explanation:

For the reaction  

Kp is defined as:

The conditions in the system are:

                   O2      N2             NO

initial            5 atm     8 atm           0

equilibrium     5-x           8-x              2x

From the stoichiometric relationship in the reaction we know that to produce 2x of NO we need x of N2 and O2.

Replacing the values in the expression for Kp we get

working with this expression

This a quadratic equation with

a=(4-Kp)

b=13*Kp

C=-40*Kp

Replacing the given value for Kp we get:

a=3.9975

b=0.0325

C=-0.1

the solution for this equation is given by the next equation:

we will get to values for X

Since negative partial pressures don't have physical meaning the solution for the system is x=0.1541.

So the partial pressure of nitrogen at the equilibrium is

i don t now

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.