\(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)
Consider the above reaction, the limiting reagent of the reaction and number of moles of NH3 formed, respectively are :
Consider the above reaction, the limiting reagent of the reaction and number of moles of NH3 formed, respectively are :
H2, 1.42 moles
H2, 0.71 moles
N2, 1.42 moles
N2, 0.71 moles
The Correct Option is C
Solution and Explanation
\(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)
28 g N2 reacts with 6 g H2 limiting reagent is N2
∴ Amount of NH3 formed on reacting 20 g N2 is,
\(=\frac{34 \times 20}{28}\)
\(=24.28\) g
\(= 1.42\) moles
So, the correct option is (C): N2, 1.42 moles
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Concepts Used:
Law of Chemical Equilibrium
Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.
Let us consider a general reversible reaction;
A+B ↔ C+D
After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.
Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.
By applying the Law of Mass Action;
The rate of forward reaction;
Rf = Kf [A]a [B]b
The rate of backward reaction;
Rb = Kb [C]c [D]d
Where,
[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.
a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.
Kf and Kb are the rate constants of forward and backward reactions.
However, at equilibrium,
Rate of forward reaction = Rate of backward reaction.
Kc is called the equilibrium constant expressed in terms of molar concentrations.
The above equation is known as the equation of Law of Chemical Equilibrium.