Question

# If you mix 2.00 moles of H2, 1.00 moles of N2 and 2.00 moles of NH3...

If you mix 2.00 moles of H2, 1.00 moles of N2 and 2.00 moles of NH3 in a one liter vessel at 472 C. Will H2 and N2 react to form NH3?

As we know that according to the Haber's process of ammonia (NH3), ammonia (NH3) is formed as follows:-

N2(g)   + 3H2(g) 2NH3(g)

1 mol 3 mol 2 mol

suppose above mixture present in one liter vessel then

molar concentration of N2 i.e [ N2 ] = no. of moles of N2 / volume of vessel = 1 / 1 = 1 M

molar concentration of H2 i.e [ H2 ] = no. of moles of H2 / volume of vessel = 3 / 1 = 3 M

molar concentration of NH3 i.e [ NH3 ] = no. of moles of NH3 / volume of vessel = 2 / 1 = 2 M

Equilibrium constant of reaction as follows:-

Equilibrium constant (Keq) = [ NH3 ]2 / [ N2 ][ H2 ]3

Equilibrium constant (Keq) = [ 2]2 / [ 1][ 3]3

Equilibrium constant (Keq) = 2 2 /1 3 3 3

Equilibrium constant (Keq) = 4 /27

Equilibrium constant (Keq) = 0.148

Given:-

volume of vessel = 1 liter

molar concentration of N2 i.e [ N2 ] = no. of moles of N2 / volume of vessel = 1 / 1 = 1 M

molar concentration of H2 i.e [ H2 ] = no. of moles of H2 / volume of vessel = 2 / 1 = 2 M

molar concentration of NH3 i.e [ NH3 ] = no. of moles of NH3 / volume of vessel = 2 / 1 = 2 M

as we know that

N2(g)   + 3H2(g) 2NH3(g)

1 mol 2 mol 2 mol

then  Equilibrium constant of reaction as follows:-

Equilibrium constant (K'eq) = [ NH3 ]2 / [ N2 ][ H2 ]3

Equilibrium constant (K'eq) =   [ 2]2 /[ 1][ 2]3

Equilibrium constant (K'eq) = 2 2 / 1 2 2 2

Equilibrium constant (K'eq) = 4 / 8

Equilibrium constant (K'eq) = 0.50

As we know that

Equilibrium constant (K'eq) = [ NH3 ]2 / [ N2 ][ H2 ]3

If the value of Equilibrium constant (Keq) increases then its indicates that the increase the formation of NH3  i.e N2 and

H2 reacts with each other to form NH3 in other words Value of Equilibrium constant (Keq) increase with decrease the concentration of  N2 and H2 and vice versa.

Since the value of Equilibrium constant (K'eq) of the given reaction i.e K'eq = 0.50 is increased as compared to the value of Equilibrium constant (Keq) of the haber's process of the formation of ammonia i.e Keq = 0.148.

As we know that

-Since the value of equilibrium constant (Keq) is independent of the original concentration of the reactants because it has a definite value for every reaction at a given temperature.

- Value of equilibrium constant (Keq) telles us the extent to which the forward or backward reaction has taken place. greater value of equilibrium constant (Keq) means that the reaction proceeded to a greater extent in the forword direction i.e the formation of product (NH3 ).

- As we know that the formation of ammonia (NH3 ) through haber's process is exothermic reaction therefore decrease the temperature of the reaction which leads the formation of ammonia (NH3 ) according to the Le-Chatelier's principle.

The answer is H2 and N2 will react to form NH3 when we mixed 2.00 moles of H2, 1.00 moles of N2 and 2.00 moles of NH3 in a one liter vessel at 472 C.

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