Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0015 M•s '; 2 NH, (g) → N, (g) + 3 H, (g) Suppose a 500. mL flask is charged under these conditions with 150. mmol of ammonia. After how much time is there only 75.0 mmol left? 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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Using a zero order integrated rate law to find concentration...
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- 1
Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0015 M's
:
2 NH, (g) - N, (g) + 3 H, (g)
2 ΝΗ,
-N2 (g)+3 H, (g)
Suppose a 500. mL flask is charged under these conditions with 150. mmol of ammonia. After how much time is there only 75.0 mmol left? 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.
x10
||
Transcribed Image Text:Using a zero order integrated rate law to find concentration... 3/5 - 1 Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0015 M's : 2 NH, (g) - N, (g) + 3 H, (g) 2 ΝΗ, -N2 (g)+3 H, (g) Suppose a 500. mL flask is charged under these conditions with 150. mmol of ammonia. After how much time is there only 75.0 mmol left? 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. x10 ||
Expert Solution
Step 1

The given reaction is :

2NH3(g)    N2(g)  +  3H2(g)

Rate constant = 0.0015 M.s-1

Volume of the flask = 500 mL

Number of millimoles of ammonia = 150 mmol

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