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So initially they are mixed with the mole ratio of $\ce{H2}$ and $\ce{N2}$ being 3:1 respectively. It seems as the reaction proceeds, both these quantities decrease and the quantity of ammonia increases until a point at which $\ce{NH3}$ stagnates. What happens at this point?

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In the graph the following reaction will occur: $$\ce{N2 +3H2 <=> 2NH3}$$ This reaction is an equilibrium reaction and you can see that there is exponential decay for the reactants while the product is formed. After a while the concentrations do not change any more, so the system is in equilibrium. This is at the end of the time span "a".

At the beginning of time span "b" the system is disturbed, as the ammonia is completely removed. The system is changed and the equilibrium has to form again. Hence the concentration of the reactants is further reduced while new product is formed. Like santiago already pointed out in the comments, this is known as Le Châtelier's principle. At the end of time span "b", the system is in equilibrium again.

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The first step to this question is writing the chemical equation for this reaction: $$\ce{N2(g) + 3H2(g) \rightleftharpoons2NH3(g)}$$ As you see, this reaction is an equilibrium reaction, meaning that not all the products will form reactants, but rather react until they reach a certain ratio. This ratio is known as the equilibrium constant, K. The equilibrium constant is given by: $$K = \frac{[A]^a[B]^b}{[C]^c[D]^d}$$ [A] and [B] are the concentrations of the products at equilibrium while [C] and [D] are the concentrations of the reactants at equilibrium. a, b, c, and d are the stoichiometric co-coefficients of their respective letter. The equilibrium constant is the same value for a particular reaction at a constant temperature and doesn't depend on the initial concentration of the products or reactants. In this case, the equilibrium constant is given by: $$K = \frac{[NH_3]^2}{[N_2][H_2]^3}$$ One more equation you need to know is the reaction quotient which gives the ratio of products to reactants at a certain time during a reaction and is given by: $$Q = \frac{[A]^a[B]^b}{[C]^c[D]^d}$$ Note how this is similar to the equilibrium constant, but here, [A], [B], [C] and [D] are the concentration of the substance at a specific time during the reaction, not at equilibrium point. This means that if Q is smaller than K, the reactants will react to form more products so that Q will increase so that it will eventually equal K and when Q is larger than K, vice versa. When Q equals K, the system is at equilibrium.

So now lets apply this concept to this graph. Initially, there is no ammonia but there is hydrogen and nitrogen gas present. This means that Q = 0 which is smaller than K as K is non-zero. This means that nitrogen and hydrogen gas will react to form ammonia. At the time corresponding to dashed line, the system is at equilibrium as the concentrations of the substances aren't reacting. This means that Q = K. Now, the sudden drop in concentration of ammonia indicates that some of the ammonia is taken out from the system. This means that Q will become smaller and hence, be smaller than K. Therefore the reactants, hydrogen and nitrogen gas, must react further to form more ammonia until reaches equilibrium again.

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