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CHXII04:CHEMICAL KINETICS

320201 $2 N_{2} O_{5} (g) \to 4 {NO}_{2} (g) + O_{2} (g)$ . The rate law for the above reaction is

1 rate

= k {[N_{2} O_{5}]}^{2}

2 rate

= {[N_{2} O_{5}]}^{2}

3 rate

= \frac{k {[N_{2} O_{5}]}^{2}}{{[{NO}_{2}]}^{4} [O_{2}]}

4 rate

= k {[N_{2} O_{5}]}^{x}

CHXII04:CHEMICAL KINETICS

320202 For the reaction, $A \to$ products, $- \frac{d [A]}{dt} = k$ and at different time interval, [A] values are
supporting img

At 20 minute, rate will be

1

12 mol / \min

2

10 mol / \min

3

8 mol / \min

4

0.4 mol / \min

Explanation:

At 20 min , moles of 'A' will be 12 mol ,
So, rate $= \frac{(20 - 12) mol}{20 \min} = 0 \cdot 4 mol / \min$

CHXII04:CHEMICAL KINETICS

320203 The rate expression for the reaction $A (g) + B (g) \to C (g) i s {r a t e = K C}_{A}^{2} C_{B}^{1 / 2}$ What changes in the initial concentration of A and B will cause the rate of reaction increase by a factor of eight?

1

C_{A} \times 2; C_{B} \times 2

2

C_{A} \times 2; C_{B} \times 4

3

C_{A} \times 1; C_{B} \times 4

4

C_{A} \times 4; C_{B} \times 1

Explanation:

Doubling the concentration of A and 4 times of the concentration of $B$ , rate increases by 8 .

CHXII04:CHEMICAL KINETICS

320204 $aA + bB \to$ Product, $dx / dt = k [A]^{a} [B]^{b}$ . If conc. of A is doubled, rate becomes four times. If conc. of B is made four times, rate is doubled. What is the relation between rate of disappearance of A and that of B ?

1

{- \frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

2

{- \frac{d [A]}{dt}} = 4 {- \frac{d [B]}{dt}}

3

- 4 {\frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

4

- 2 {\frac{d [A]}{dt}} = - \frac{1}{2} {\frac{d [B]}{dt}}

Explanation:

Rate $= \frac{d x}{d t} = K [A]^{a} [B]^{b}$
If [A] is doubled, rate becomes 4 times, means second order w.r.t. A.
If [B] is made 4 times, rate becomes 2 times,means $\frac{1}{2}$ order w.r.t. B.
$\Rightarrow 2 A + \frac{1}{2} B \to$ Product
$\Rightarrow - \frac{1}{2} \frac{d [A]}{dt} = \frac{- 2 d [B]}{dt}$
$\Rightarrow - \frac{d [A]}{dt} = 4 (- \frac{d [B]}{dt})$

CHXII04:CHEMICAL KINETICS

320205 The rate law for a reaction between the substances A and B is given by
$r a t e = k [A]^{n} {[B]}^{m}$
On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

1

2^{(n - m)}

2

1 / 2^{(m + n)}

3

(m + n)

4

(n - m)

Explanation:

Given, $(rate)_{1} = k [A]^{n} {[\frac{B}{2}]}^{m}$
On doubling the concentration of A and halving the concentration of $B$ , the rate can be represented as, $(rate)_{2} = k [2 A]^{n} [B / 2]^{m}$
$\frac{(rate)_{2}}{(rate)_{1}} = \frac{k [2 A]^{n} {[\frac{B}{2}]}^{m}}{k [A]^{n} [B]^{m}} = 2^{n - m}$

CHXII04:CHEMICAL KINETICS

320201 $2 N_{2} O_{5} (g) \to 4 {NO}_{2} (g) + O_{2} (g)$ . The rate law for the above reaction is

1 rate

= k {[N_{2} O_{5}]}^{2}

2 rate

= {[N_{2} O_{5}]}^{2}

3 rate

= \frac{k {[N_{2} O_{5}]}^{2}}{{[{NO}_{2}]}^{4} [O_{2}]}

4 rate

= k {[N_{2} O_{5}]}^{x}

CHXII04:CHEMICAL KINETICS

320202 For the reaction, $A \to$ products, $- \frac{d [A]}{dt} = k$ and at different time interval, [A] values are
supporting img

At 20 minute, rate will be

1

12 mol / \min

2

10 mol / \min

3

8 mol / \min

4

0.4 mol / \min

Explanation:

At 20 min , moles of 'A' will be 12 mol ,
So, rate $= \frac{(20 - 12) mol}{20 \min} = 0 \cdot 4 mol / \min$

CHXII04:CHEMICAL KINETICS

320203 The rate expression for the reaction $A (g) + B (g) \to C (g) i s {r a t e = K C}_{A}^{2} C_{B}^{1 / 2}$ What changes in the initial concentration of A and B will cause the rate of reaction increase by a factor of eight?

1

C_{A} \times 2; C_{B} \times 2

2

C_{A} \times 2; C_{B} \times 4

3

C_{A} \times 1; C_{B} \times 4

4

C_{A} \times 4; C_{B} \times 1

Explanation:

Doubling the concentration of A and 4 times of the concentration of $B$ , rate increases by 8 .

CHXII04:CHEMICAL KINETICS

320204 $aA + bB \to$ Product, $dx / dt = k [A]^{a} [B]^{b}$ . If conc. of A is doubled, rate becomes four times. If conc. of B is made four times, rate is doubled. What is the relation between rate of disappearance of A and that of B ?

1

{- \frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

2

{- \frac{d [A]}{dt}} = 4 {- \frac{d [B]}{dt}}

3

- 4 {\frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

4

- 2 {\frac{d [A]}{dt}} = - \frac{1}{2} {\frac{d [B]}{dt}}

Explanation:

Rate $= \frac{d x}{d t} = K [A]^{a} [B]^{b}$
If [A] is doubled, rate becomes 4 times, means second order w.r.t. A.
If [B] is made 4 times, rate becomes 2 times,means $\frac{1}{2}$ order w.r.t. B.
$\Rightarrow 2 A + \frac{1}{2} B \to$ Product
$\Rightarrow - \frac{1}{2} \frac{d [A]}{dt} = \frac{- 2 d [B]}{dt}$
$\Rightarrow - \frac{d [A]}{dt} = 4 (- \frac{d [B]}{dt})$

CHXII04:CHEMICAL KINETICS

320205 The rate law for a reaction between the substances A and B is given by
$r a t e = k [A]^{n} {[B]}^{m}$
On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

1

2^{(n - m)}

2

1 / 2^{(m + n)}

3

(m + n)

4

(n - m)

Explanation:

Given, $(rate)_{1} = k [A]^{n} {[\frac{B}{2}]}^{m}$
On doubling the concentration of A and halving the concentration of $B$ , the rate can be represented as, $(rate)_{2} = k [2 A]^{n} [B / 2]^{m}$
$\frac{(rate)_{2}}{(rate)_{1}} = \frac{k [2 A]^{n} {[\frac{B}{2}]}^{m}}{k [A]^{n} [B]^{m}} = 2^{n - m}$

CHXII04:CHEMICAL KINETICS

320201 $2 N_{2} O_{5} (g) \to 4 {NO}_{2} (g) + O_{2} (g)$ . The rate law for the above reaction is

1 rate

= k {[N_{2} O_{5}]}^{2}

2 rate

= {[N_{2} O_{5}]}^{2}

3 rate

= \frac{k {[N_{2} O_{5}]}^{2}}{{[{NO}_{2}]}^{4} [O_{2}]}

4 rate

= k {[N_{2} O_{5}]}^{x}

CHXII04:CHEMICAL KINETICS

320202 For the reaction, $A \to$ products, $- \frac{d [A]}{dt} = k$ and at different time interval, [A] values are
supporting img

At 20 minute, rate will be

1

12 mol / \min

2

10 mol / \min

3

8 mol / \min

4

0.4 mol / \min

Explanation:

At 20 min , moles of 'A' will be 12 mol ,
So, rate $= \frac{(20 - 12) mol}{20 \min} = 0 \cdot 4 mol / \min$

CHXII04:CHEMICAL KINETICS

320203 The rate expression for the reaction $A (g) + B (g) \to C (g) i s {r a t e = K C}_{A}^{2} C_{B}^{1 / 2}$ What changes in the initial concentration of A and B will cause the rate of reaction increase by a factor of eight?

1

C_{A} \times 2; C_{B} \times 2

2

C_{A} \times 2; C_{B} \times 4

3

C_{A} \times 1; C_{B} \times 4

4

C_{A} \times 4; C_{B} \times 1

Explanation:

Doubling the concentration of A and 4 times of the concentration of $B$ , rate increases by 8 .

CHXII04:CHEMICAL KINETICS

320204 $aA + bB \to$ Product, $dx / dt = k [A]^{a} [B]^{b}$ . If conc. of A is doubled, rate becomes four times. If conc. of B is made four times, rate is doubled. What is the relation between rate of disappearance of A and that of B ?

1

{- \frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

2

{- \frac{d [A]}{dt}} = 4 {- \frac{d [B]}{dt}}

3

- 4 {\frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

4

- 2 {\frac{d [A]}{dt}} = - \frac{1}{2} {\frac{d [B]}{dt}}

Explanation:

Rate $= \frac{d x}{d t} = K [A]^{a} [B]^{b}$
If [A] is doubled, rate becomes 4 times, means second order w.r.t. A.
If [B] is made 4 times, rate becomes 2 times,means $\frac{1}{2}$ order w.r.t. B.
$\Rightarrow 2 A + \frac{1}{2} B \to$ Product
$\Rightarrow - \frac{1}{2} \frac{d [A]}{dt} = \frac{- 2 d [B]}{dt}$
$\Rightarrow - \frac{d [A]}{dt} = 4 (- \frac{d [B]}{dt})$

CHXII04:CHEMICAL KINETICS

320205 The rate law for a reaction between the substances A and B is given by
$r a t e = k [A]^{n} {[B]}^{m}$
On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

1

2^{(n - m)}

2

1 / 2^{(m + n)}

3

(m + n)

4

(n - m)

Explanation:

Given, $(rate)_{1} = k [A]^{n} {[\frac{B}{2}]}^{m}$
On doubling the concentration of A and halving the concentration of $B$ , the rate can be represented as, $(rate)_{2} = k [2 A]^{n} [B / 2]^{m}$
$\frac{(rate)_{2}}{(rate)_{1}} = \frac{k [2 A]^{n} {[\frac{B}{2}]}^{m}}{k [A]^{n} [B]^{m}} = 2^{n - m}$

CHXII04:CHEMICAL KINETICS

320201 $2 N_{2} O_{5} (g) \to 4 {NO}_{2} (g) + O_{2} (g)$ . The rate law for the above reaction is

1 rate

= k {[N_{2} O_{5}]}^{2}

2 rate

= {[N_{2} O_{5}]}^{2}

3 rate

= \frac{k {[N_{2} O_{5}]}^{2}}{{[{NO}_{2}]}^{4} [O_{2}]}

4 rate

= k {[N_{2} O_{5}]}^{x}

CHXII04:CHEMICAL KINETICS

320202 For the reaction, $A \to$ products, $- \frac{d [A]}{dt} = k$ and at different time interval, [A] values are
supporting img

At 20 minute, rate will be

1

12 mol / \min

2

10 mol / \min

3

8 mol / \min

4

0.4 mol / \min

Explanation:

At 20 min , moles of 'A' will be 12 mol ,
So, rate $= \frac{(20 - 12) mol}{20 \min} = 0 \cdot 4 mol / \min$

CHXII04:CHEMICAL KINETICS

320203 The rate expression for the reaction $A (g) + B (g) \to C (g) i s {r a t e = K C}_{A}^{2} C_{B}^{1 / 2}$ What changes in the initial concentration of A and B will cause the rate of reaction increase by a factor of eight?

1

C_{A} \times 2; C_{B} \times 2

2

C_{A} \times 2; C_{B} \times 4

3

C_{A} \times 1; C_{B} \times 4

4

C_{A} \times 4; C_{B} \times 1

Explanation:

Doubling the concentration of A and 4 times of the concentration of $B$ , rate increases by 8 .

CHXII04:CHEMICAL KINETICS

320204 $aA + bB \to$ Product, $dx / dt = k [A]^{a} [B]^{b}$ . If conc. of A is doubled, rate becomes four times. If conc. of B is made four times, rate is doubled. What is the relation between rate of disappearance of A and that of B ?

1

{- \frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

2

{- \frac{d [A]}{dt}} = 4 {- \frac{d [B]}{dt}}

3

- 4 {\frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

4

- 2 {\frac{d [A]}{dt}} = - \frac{1}{2} {\frac{d [B]}{dt}}

Explanation:

Rate $= \frac{d x}{d t} = K [A]^{a} [B]^{b}$
If [A] is doubled, rate becomes 4 times, means second order w.r.t. A.
If [B] is made 4 times, rate becomes 2 times,means $\frac{1}{2}$ order w.r.t. B.
$\Rightarrow 2 A + \frac{1}{2} B \to$ Product
$\Rightarrow - \frac{1}{2} \frac{d [A]}{dt} = \frac{- 2 d [B]}{dt}$
$\Rightarrow - \frac{d [A]}{dt} = 4 (- \frac{d [B]}{dt})$

CHXII04:CHEMICAL KINETICS

320205 The rate law for a reaction between the substances A and B is given by
$r a t e = k [A]^{n} {[B]}^{m}$
On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

1

2^{(n - m)}

2

1 / 2^{(m + n)}

3

(m + n)

4

(n - m)

Explanation:

Given, $(rate)_{1} = k [A]^{n} {[\frac{B}{2}]}^{m}$
On doubling the concentration of A and halving the concentration of $B$ , the rate can be represented as, $(rate)_{2} = k [2 A]^{n} [B / 2]^{m}$
$\frac{(rate)_{2}}{(rate)_{1}} = \frac{k [2 A]^{n} {[\frac{B}{2}]}^{m}}{k [A]^{n} [B]^{m}} = 2^{n - m}$

CHXII04:CHEMICAL KINETICS

320201 $2 N_{2} O_{5} (g) \to 4 {NO}_{2} (g) + O_{2} (g)$ . The rate law for the above reaction is

1 rate

= k {[N_{2} O_{5}]}^{2}

2 rate

= {[N_{2} O_{5}]}^{2}

3 rate

= \frac{k {[N_{2} O_{5}]}^{2}}{{[{NO}_{2}]}^{4} [O_{2}]}

4 rate

= k {[N_{2} O_{5}]}^{x}

CHXII04:CHEMICAL KINETICS

320202 For the reaction, $A \to$ products, $- \frac{d [A]}{dt} = k$ and at different time interval, [A] values are
supporting img

At 20 minute, rate will be

1

12 mol / \min

2

10 mol / \min

3

8 mol / \min

4

0.4 mol / \min

Explanation:

At 20 min , moles of 'A' will be 12 mol ,
So, rate $= \frac{(20 - 12) mol}{20 \min} = 0 \cdot 4 mol / \min$

CHXII04:CHEMICAL KINETICS

320203 The rate expression for the reaction $A (g) + B (g) \to C (g) i s {r a t e = K C}_{A}^{2} C_{B}^{1 / 2}$ What changes in the initial concentration of A and B will cause the rate of reaction increase by a factor of eight?

1

C_{A} \times 2; C_{B} \times 2

2

C_{A} \times 2; C_{B} \times 4

3

C_{A} \times 1; C_{B} \times 4

4

C_{A} \times 4; C_{B} \times 1

Explanation:

Doubling the concentration of A and 4 times of the concentration of $B$ , rate increases by 8 .

CHXII04:CHEMICAL KINETICS

320204 $aA + bB \to$ Product, $dx / dt = k [A]^{a} [B]^{b}$ . If conc. of A is doubled, rate becomes four times. If conc. of B is made four times, rate is doubled. What is the relation between rate of disappearance of A and that of B ?

1

{- \frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

2

{- \frac{d [A]}{dt}} = 4 {- \frac{d [B]}{dt}}

3

- 4 {\frac{d [A]}{dt}} = {- \frac{d [B]}{dt}}

4

- 2 {\frac{d [A]}{dt}} = - \frac{1}{2} {\frac{d [B]}{dt}}

Explanation:

Rate $= \frac{d x}{d t} = K [A]^{a} [B]^{b}$
If [A] is doubled, rate becomes 4 times, means second order w.r.t. A.
If [B] is made 4 times, rate becomes 2 times,means $\frac{1}{2}$ order w.r.t. B.
$\Rightarrow 2 A + \frac{1}{2} B \to$ Product
$\Rightarrow - \frac{1}{2} \frac{d [A]}{dt} = \frac{- 2 d [B]}{dt}$
$\Rightarrow - \frac{d [A]}{dt} = 4 (- \frac{d [B]}{dt})$

CHXII04:CHEMICAL KINETICS

320205 The rate law for a reaction between the substances A and B is given by
$r a t e = k [A]^{n} {[B]}^{m}$
On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

1

2^{(n - m)}

2

1 / 2^{(m + n)}

3

(m + n)

4

(n - m)

Explanation:

Given, $(rate)_{1} = k [A]^{n} {[\frac{B}{2}]}^{m}$
On doubling the concentration of A and halving the concentration of $B$ , the rate can be represented as, $(rate)_{2} = k [2 A]^{n} [B / 2]^{m}$
$\frac{(rate)_{2}}{(rate)_{1}} = \frac{k [2 A]^{n} {[\frac{B}{2}]}^{m}}{k [A]^{n} [B]^{m}} = 2^{n - m}$

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