1)The reaction shownis exothermic. If the reaction at equilibrium is cooled, whi

Question

1)The reaction shownis exothermic. If the reaction at equilibrium is cooled, which of the following will be true?

N2(g) + O2(g) <====> 2 NO(g)

Keq = 5 x 1031

the position of equilibrium will shift toward the NO, and the value of Keq will increase

the position of equilibrium will shift toward the NO, and the value of Keq will decrease

the position of equilibrium will shift toward the N2 and O2, and the value of Keq will increase

the position of equilibrium will shift toward the N2 and O2, and the value of Keq will decrease

the position of equilibrium will shift toward the NO, and the value of Keq will remain the same

2)Which material in the following mechanism is a catalyst?

Step 1

A + B3 AB + B2

slow

Step 2

AB A + B

fast

Step 3

B + B3 2 B2

fast

A

AB

B

B3

B2

3) Which material in the following mechanism is an intermediate?

Step 1

A + B3 AB + B2

slow

Step 2

AB A + B

fast

Step 3

B + B3 2 B2

fast

A

B3

B

B2

none of the above

4) The activation energy was measured for both the forward (Ea=150 kJ/molrxn) and reverse (Ea = 95 kJ/molrxn) directions of a reversible reaction. What would be the activation energy for the reverse reaction in the presence of a catalyst that decreased the activation energy for the forward reaction to 125 kJ/ molrxn?

25 kJ/molrxn

75 kJ/ molrxn

120 kJ/ molrxn

30 kJ/ molrxn

none of these

5)

The initial rate of disappearance of bromine (Br2) for the reaction shown below was measured for several different concentrations of bromine, CH3COCH3, and H+ ions. (Note that H+ is a catalyst in this reaction – it participates in the reaction but is not itself consumed) CH3COCH3 + Br2 CH3COCH2Br + Br   Which of the following statements about the orders is correct?

Trial Number -

Initial [CH3COCH3]

- Initial [Br2]-

Initial [H+]-

Initial Reaction Rate, Ms1

1

0.30

0.050

0.050

6.0 x 10–5

2

0.30

0.10

0.050

6.0 x 10–5

3

0.30

0.050

0.10

1.2 x 10–4

4

0.40

0.050

0.20

3.2 x 10–4

zero-order in H+ and first-order in Br2

first-order in H+ and first-order in Br2

first-order in H+ and zero-order in Br2

second-order in H+ and zero-order in Br2

none of the above.

6)If a reaction is zero order with respect to a reactant, when the concentration of that reaction is changed to 5 times larger than its original, and all the other factors are kept constant, the rate of the reaction will be

5 times faster than the original rate

5 times slower than the original rate

10 times faster than the original rate

25 times faster than the original rate

the same

N2(g) + O2(g) <====> 2 NO(g)

Keq = 5 x 1031

Explanation / Answer

1) given the reaction is exothermic

so

the reaction can be written as

N2 + 02 ---> 2N0 + dH

so

if the reaction is cooled

it means that heat is removed from the reaction

so

according to Le chatlier priniciple

the reaction will proceed in a direction to produce more heat

so

the equilibrium will shift towards the formation of N0

also

according to arhenius equation

ln( K2 /K1 ) = ( dH / R ) ( 1/ T1 - 1/T2)

given

the temperatue is cooled

so

T2 < T1

1/T2 > 1/T1

1/T1 - 1/T2 < 1

also

for exothermic reaction

dH = -ve

so

ln ( K2 / K1) = +ve

so

k2 > k1

so

Keq will increase

so the answer is

   

the position of equilibrium will shift toward the NO, and the value of Keq will increase

2)

we know that

a catalyst reacts in the beginning and reappears later

here

A reacts in step1 and then reappears in step2

so

here

the catalyst is A

3)

an intermediate will not be present in the beginning. It forms in the middle and then disappers later

here

AB is formed in the step1 and diappears in step2

so

the intermediate is AB

4)

the activation energy of the reverse reaction should be lowered by the same amount

so

the Ea for reverse reaction should be 95-25 = 70

so

the answer is none of these

5)

Let the rate law be

rate = K [CH3COCH3]^a [Br2]^b [H+]^c

now

consider trail 1 and trail 2

we see that

[CH3COCH3] and [H+] are constant

so

the equation becomes

rate2/ rate1 = [[Br2]2 / [Br2]1]^b

6 x 10-5 / 6 x 10-5 = [0.1/0.05]^b

1 = [2]^b

so

b = 0

so

the order with respect to Br2 is zero

now consider trai 1 and trail 3

rate3 / rate1 = ( [H+]3/ [H+]1 )^c

1.2 x 10-4 / 6 x 10-5 = ( 0.1 / 0.05)^c

2 = (2)^c

c = 1

so

it is first order with respect to [H+]

so

the answer is

first-order in H+ and zero-order in Br2

6)

given the order is zero

so

it becomes

rate = K [A]^0

rate = K

so rate does not depend on reactant A

so

it remains same

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