Question

In the lab activity, the reaction rate was determined by the appearance of a product. However, the reaction rate can also be determined by the disappearance of a reactant. Rate =Δ[product]/Δt or rate−Δ[reactant]Δt. In each situation below, you are given a rate measured by the appearance of one component of the reaction and are asked to predict the rate of appearance or disappearance of another component, based on logic and stoichiometric relationships.
For example, if the reaction is as follows:

A+2B⟶products

For every mole of A that is used, 2 moles of B are used so the rate of disappearance of B is twice the rate of the disappearance of A.

This may be expressed as:

rate=−Δ[B]/Δt=−2[A]/Δt , N2(g)+3H2(g)⟶2NH3(g)

The reaction rate is measured as 0.032 M NH3/s. Determine the rate of disappearance of N2 and the rate of disappearance of H2. Explain how you arrived at your answers.

Answer 1

Answer: Rate of disappearance of $N_2$= 0.032 M/s

Rate of disappearance of $H_2$ = 0.096 M/s

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of $N_2$ = $-\frac{1d[N_2]}{dt}$

Rate in terms of disappearance of $H_2$ = $-\frac{1d[H_2]}{3dt}$

Rate in terms of appearance of $NH_3$ = $\frac{1d[NH_3]}{2dt}$

Rate = $-\frac{1d[N_2]}{dt}=-\frac{1d[H_2]}{3dt}=\frac{1d[NH_3]}{2dt}$

Given : = 0.032 M/s

Rate of disappearance of =

Rate of disappearance of $H_2$ = $-\frac{d[H_2]}{dt}=3\times 0.032M/s=0.096M/s$