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?
Answer:-
As we know that according to the Haber's process of ammonia (NH_{3}), ammonia (NH_{3}) is formed as follows:-
N_{2(g) } + 3H_{2(g)} 2NH_{3(g)}
1 mol 3 mol 2 mol
suppose above mixture present in one liter vessel then
molar concentration of N_{2} i.e [ N_{2} ] = no. of moles of N_{2} / volume of vessel = 1 / 1 = 1 M
molar concentration of H_{2} i.e [ H_{2} ] = no. of moles of H_{2} / volume of vessel = 3 / 1 = 3 M
molar concentration of NH_{3} i.e [ NH_{3} ] = no. of moles of NH_{3} / volume of vessel = 2 / 1 = 2 M
Equilibrium constant of reaction as follows:-
Equilibrium constant (Keq) = [ NH_{3} ]^{2} / [ N_{2} ][ H_{2} ]^{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 N_{2} i.e [ N_{2} ] = no. of moles of N_{2} / volume of vessel = 1 / 1 = 1 M
molar concentration of H_{2} i.e [ H_{2} ] = no. of moles of H_{2} / volume of vessel = 2 / 1 = 2 M
molar concentration of NH_{3} i.e [ NH_{3} ] = no. of moles of NH_{3} / volume of vessel = 2 / 1 = 2 M
as we know that
N_{2(g) } + 3H_{2(g)} 2NH_{3(g)}
1 mol 2 mol 2 mol
then Equilibrium constant of reaction as follows:-
Equilibrium constant (K'eq) = [ NH_{3} ]^{2} / [ N_{2} ][ H_{2} ]^{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) = [ NH_{3} ]^{2} / [ N_{2} ][ H_{2} ]^{3}
If the value of Equilibrium constant (Keq) increases then its indicates that the increase the formation of NH_{3 }i.e N_{2} and
H_{2} reacts with each other to form NH_{3} in other words Value of Equilibrium constant (Keq) increase with decrease the concentration of N_{2} and H_{2} 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 (NH_{3} ).
- As we know that the formation of ammonia (NH_{3} ) through haber's process is exothermic reaction therefore decrease the temperature of the reaction which leads the formation of ammonia (NH_{3} ) according to the Le-Chatelier's principle.
The answer is H_{2} and N_{2} will react to form NH3 when we mixed 2.00 moles of H_{2}, 1.00 moles of N_{2} and 2.00 moles of NH_{3} in a one liter vessel at 472 C.
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