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drek231 [11]
3 years ago
15

Consider this reaction:2HI(g) ------> H2(g) + I2(g)At a certain temperature it obeys this rate law.rate = (8.74 x 10^-4s^-1)

[HI] 1. Suppose a vessel contains HI at a concentration of 0.330M . Calculate the concentration of HI in the vessel 800 seconds later. You may assume no other reaction is important.
Chemistry
1 answer:
nasty-shy [4]3 years ago
8 0

Answer:

0.1440M

Explanation:

Let''s bring out the parameters we were given;

Rate constant = 8.74 x 10^-4s^-1

Initial Concentration [A]o = 0.330M

Final concentration [A]= ?

Time = 800s

The reaction is a first order reaction, due to the unit of the rate constant. In first order reactions, the reaction rate is directly proportional to the reactant concentration and the units of first order rate constants are 1/sec.

Formular relating these parameters is given as;

ln[A] = ln[A]o − kt

Making [A] subject of interest, we have;

ln[A] = ln[A]o − kt

ln[A]  = ln(0.330) - ( 8.74 x 10^-4 * 800)

In[A] = - 1.1086 - (6992 x 10^-4)

ln[A] = -1.8079

[A] = 0.1440M

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Answer:

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Explanation:

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Consider an acid in a form HA (aq) and base in a form of B (aq). Since acid is a proton donor, it will donate its hydrogen ion to the base, B. The resultant products would be A^{-} (aq) and BH^{+} (aq).

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Answer:

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Explanation:

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