Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0089 M. s^-1:

2NH3 (g) → N2 (g) + 3H2 (g)

Suppose a 5.0 L flask is charged under these conditions with 400. mmol of ammonia. How much is left 2.0 s later? You may assume no other reaction is important.

Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

Question

Mike Adams

Chemistry

28 May, 21:12

Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0089 M. s^-1:

2NH3 (g) → N2 (g) + 3H2 (g)

Suppose a 5.0 L flask is charged under these conditions with 400. mmol of ammonia. How much is left 2.0 s later? You may assume no other reaction is important.

Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

+4

Answers (1)

Know the Answer?

Amara Marsh · Answer 1

The rate of reaction will not depend upon concentration of reactant. It will be always constant and equal to. 0089M s⁻¹.

Initial moles of reactant = 400 x 10⁻³ mole in 5 L

molarity = 400 x 10⁻³ / 5 M

= 80 x 10⁻³ M.

=.08M

no of moles reacted in 2 s =.0089 x 2

=.0178 M

concentration left =.08 -.0178 M

=.0622 M.

No of moles left in 5 L

= 5 x. 0622 =.31 moles.