Consider the reaction:
N2(g) + 3 H2(g) --> 2 NH3(g).
At a particular time, the nitrogen is being consumed at a rate of 0.60 moles/sec. At this same time, what is the rate at which ammonia is being formed?
a.
3.0 mol/s
b.
0.12 mol/s
c.
0.30 mol/s
d.
2.0 mol/s
e.
1.2 mol/s
This is just another stoichiometry problem.
0.6 x (2 mol NH3/1 mol N2) = ?
Better think again.
To find the rate at which ammonia (NH3) is being formed, we can use the stoichiometry of the reaction. According to the balanced equation:
N2(g) + 3 H2(g) --> 2 NH3(g)
It takes 1 mole of N2 to produce 2 moles of NH3. This means that for every mole of N2 consumed, 2 moles of NH3 are formed.
Given that the nitrogen (N2) is being consumed at a rate of 0.60 moles/sec, the rate at which ammonia is being formed will be twice that, since the stoichiometry of the reaction is 1:2.
Rate of ammonia formation = 2 * Rate of nitrogen consumption
Rate of ammonia formation = 2 * 0.60 mol/s
Rate of ammonia formation = 1.20 mol/s
Therefore, at this particular time, the rate at which ammonia is being formed is 1.20 mol/s.