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sergiy2304 [10]
4 years ago
8

By what factor does the rate change in each of the following cases (assuming constant temperature)? (a) A reaction is first orde

r in reactant A, and [A] is doubled. (b) A reaction is second order in reactant B, and [B] is halved. (c) A reaction is second order in reactant C, and [C] is tripled.
Chemistry
1 answer:
Papessa [141]4 years ago
4 0

Answer:

a) 2

b) 1/4

c) 9

Explanation:

a) for a first order reaction in reactant A

r initial = k*[A initial]

then if the concentration is doubled [A final ]= 2*[A initial]  , then

r final = k*[A final ] = 2* k*[A initial] = 2*r initial

then the velocity changes by a factor of 2

b) for a second order reaction in reactant B

r initial = k*[B initial]²

then if the concentration is halved: [B final ]= [B initial]/2  , then

r final = k*[B final ]²  = k*( [B initial]/2 )²  =k* [B initial]² /4 = r initial /4

then the velocity changes by a factor of 1/4

c) for a second order reaction in reactant C

r initial = k*[C initial]²

then if the concentration is tripled : [C final ]= 3* [C initial]  , then

r final = k*[C final ]²  = k*( 3*[C initial] )²  =k* [C initial]² *9 = 9 * r initial  

then the velocity changes by a factor of 9

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Answer : The volume of water added are, 15 mL

Explanation :

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of HCl.

M_2\text{ and }V_2 are the final molarity and volume of water.

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Answer:

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Explanation:

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For molecules b, c and e we have <u>hydrogen bond to a heteroatom</u> (O, N, S, or P). In this case oxygen, therefore we will have <u>hydrogen bonding </u>interactions (a very strong interaction). So, we can discard these ones.

In molecule e, we have "Cl" bond to a "C" therefore we will have the presence of a <u>dipole</u> (due to the <u>electronegativity difference</u>). If we have a dipole, we will have a <u>dipole-dipole interaction</u> (a strong interaction, less than hydrogen bonding but still is a strong interaction).

In molecule a, we have only <u>Van der Waals interactions</u> because in this molecule we have only carbon and hydrogen atoms bonded by single bonds. So, we will have a n<u>on-polar molecule</u>. These interactions are the weakest interactions of all the molecules given. So, <u>if we have weaker interactions the molecules can be converted to a gas state more easily and we have more vapor pressure.  </u>

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