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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.

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  1. 28 May, 21:26
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    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.
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