Question

Consider the following reaction.
N2(g) + 3H2(g) ⭢ 2NH3(g)
If the rate of loss of hydrogen gas is 0.03mol L-1 s-1, what is the rate of production of ammonia?

Answer 1

Given reaction is

N2(g) + 3H2(g) ––> 2NH3(g)

If the reaction rate = r

Then according to the above

stoichiometric equation,

Rate of reaction, r = –(d[H2]/dt)

= (d[NH3]/dt)

So,

–(1/3)(d[H2]/dt) = (1/2)(d[NH3]/dt)

Or, –(d[H2]/dt) =(3/2)(d[NH3]/dt)

______[A]

Given rate of Loss of Hydrogen gas

–(d[H2]/dt) = 0.03 mol L-1 s-1

Putting the value in equation [A]

(d[NH3]/dt) = (3/2)×0.03 mol L-1 s-1

Or, (d[NH3]/dt) = 0.045 mol L-1 s-1

So, Rate of production of ammonia

= 0.045 mol L-1 s-1