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CHXII04:CHEMICAL KINETICS

320535 Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide
$2N O_{(g)} + O_{2} (g) \overset{K}{\to} 2N O_{2(g)}$
If the mechanism of reaction is
$NO + O_{2} \underset{K}{⇌} N_{3} ( fast ) N O_{3} + NO \overset{K_{1}}{\to}$
$N O_{2} + N O_{2} (slow)$
then rate law is

1 Rate

= K^{'} [N O] [O_{2}]

2 Rate

= K^{'} [NO] {[O_{2}]}^{2}

3 Rate

= K^{'} [N O]^{2} [O_{2}]

4 Rate

= K^{'} [N O]^{3} [O_{2}]

Explanation:

From slowest step i.e., rds step,
Rate $= k_{1} [{NO}_{3}] [NO]$
${NO}_{3}$ is an intermediate, need to be replaced with $NO$ and $O_{2}$ by using fast step.
$K = \frac{[N O_{3}]}{[N O] [O_{2}]} \Rightarrow [N O_{3}] = K [N O] [O_{2}]$
Substitute $[{NO}_{3}]$ in eq. (1)
Rate $= k_{1} K [N O] [O_{2}] [N O]$
Rate $= k^{'} [N O]^{2} [O_{2}] [k^{'} = k_{1} K]$

CHXII04:CHEMICAL KINETICS

320536 The plot of $t_{1 / 2} v / s {[R]}_{0}$ for a reaction is a straight-line parallel to $x$ -axis. The unit for the rate constant of this reaction is

1

{molL}^{- 1} s

2

{molL}^{- 1} s^{- 1}

3

{Lmol}^{- 1} s^{- 1}

4

s^{- 1}

Explanation:

Plot of $t_{1 / 2} vs [R]_{0}$ for a reaction is a straight line parallel to $x$ -axis, means $t_{1 / 2} vs [R]_{0}$ is independent of $[R]_{0}$ hence, the reaction is first order.
$∴$ Unit of rate constant $= s^{- 1}$

CHXII04:CHEMICAL KINETICS

320537 The reaction, $X + 2 Y + Z \to N$ occurs by the following mechanism

(i) $X + Y ⇌ M$ very rapid equilibrium)

(ii) $M + Z \to O$ (slow) (iii) $O + Y \to N$ (very fast)

What is the rate law for this reaction?

1

R a t e = k [Z]

2

R a t e = k [X] [Y]^{2} [Z]

3

R a t e = k [N]

4

R a t e = k [X] [Y] [Z]

Explanation:

$K_{C} = \frac{[M]}{[X] [Y]} \Rightarrow [M] = K_{C} [X] [Y]$
$R a t e = k^{'} [M] [Z] = k^{'} K_{C} [X] [Y] [Z] = K [X] [Y] [Z]$

CHXII04:CHEMICAL KINETICS

320538 The reaction, $X_{2} + Y_{2} \to 2 X Y$ occurs by the following mechanism

(i) $X_{2} ⇌ 2 X$ (fast)

(ii) $X + Y_{2} \to X Y + X$ (slow)

(iii) $X + Y \to X Y$ (fast)
What is the order of the reaction?

1 2

2 0

3 -1

4 1.5

Explanation:

From slowest step (rds),
$r = k [X] [Y_{2}]$
From fast step
$K = \frac{{[X]}^{2}}{[X_{2}]}$
$[X] = K^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}}$
$∴ r = k K^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}} [Y_{2}]$
$r = k^{'} {[X_{2}]}^{\frac{1}{2}} [Y_{2}] (k^{'} = k K^{\frac{1}{2}})$
Order of the reaction $= 1 + 1 / 2 = 3 / 2 o r 1 .5$

CHXII04:CHEMICAL KINETICS

320535 Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide
$2N O_{(g)} + O_{2} (g) \overset{K}{\to} 2N O_{2(g)}$
If the mechanism of reaction is
$NO + O_{2} \underset{K}{⇌} N_{3} ( fast ) N O_{3} + NO \overset{K_{1}}{\to}$
$N O_{2} + N O_{2} (slow)$
then rate law is

1 Rate

= K^{'} [N O] [O_{2}]

2 Rate

= K^{'} [NO] {[O_{2}]}^{2}

3 Rate

= K^{'} [N O]^{2} [O_{2}]

4 Rate

= K^{'} [N O]^{3} [O_{2}]

Explanation:

From slowest step i.e., rds step,
Rate $= k_{1} [{NO}_{3}] [NO]$
${NO}_{3}$ is an intermediate, need to be replaced with $NO$ and $O_{2}$ by using fast step.
$K = \frac{[N O_{3}]}{[N O] [O_{2}]} \Rightarrow [N O_{3}] = K [N O] [O_{2}]$
Substitute $[{NO}_{3}]$ in eq. (1)
Rate $= k_{1} K [N O] [O_{2}] [N O]$
Rate $= k^{'} [N O]^{2} [O_{2}] [k^{'} = k_{1} K]$

CHXII04:CHEMICAL KINETICS

320536 The plot of $t_{1 / 2} v / s {[R]}_{0}$ for a reaction is a straight-line parallel to $x$ -axis. The unit for the rate constant of this reaction is

1

{molL}^{- 1} s

2

{molL}^{- 1} s^{- 1}

3

{Lmol}^{- 1} s^{- 1}

4

s^{- 1}

Explanation:

Plot of $t_{1 / 2} vs [R]_{0}$ for a reaction is a straight line parallel to $x$ -axis, means $t_{1 / 2} vs [R]_{0}$ is independent of $[R]_{0}$ hence, the reaction is first order.
$∴$ Unit of rate constant $= s^{- 1}$

CHXII04:CHEMICAL KINETICS

320537 The reaction, $X + 2 Y + Z \to N$ occurs by the following mechanism

(i) $X + Y ⇌ M$ very rapid equilibrium)

(ii) $M + Z \to O$ (slow) (iii) $O + Y \to N$ (very fast)

What is the rate law for this reaction?

1

R a t e = k [Z]

2

R a t e = k [X] [Y]^{2} [Z]

3

R a t e = k [N]

4

R a t e = k [X] [Y] [Z]

Explanation:

$K_{C} = \frac{[M]}{[X] [Y]} \Rightarrow [M] = K_{C} [X] [Y]$
$R a t e = k^{'} [M] [Z] = k^{'} K_{C} [X] [Y] [Z] = K [X] [Y] [Z]$

CHXII04:CHEMICAL KINETICS

320538 The reaction, $X_{2} + Y_{2} \to 2 X Y$ occurs by the following mechanism

(i) $X_{2} ⇌ 2 X$ (fast)

(ii) $X + Y_{2} \to X Y + X$ (slow)

(iii) $X + Y \to X Y$ (fast)
What is the order of the reaction?

1 2

2 0

3 -1

4 1.5

Explanation:

From slowest step (rds),
$r = k [X] [Y_{2}]$
From fast step
$K = \frac{{[X]}^{2}}{[X_{2}]}$
$[X] = K^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}}$
$∴ r = k K^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}} [Y_{2}]$
$r = k^{'} {[X_{2}]}^{\frac{1}{2}} [Y_{2}] (k^{'} = k K^{\frac{1}{2}})$
Order of the reaction $= 1 + 1 / 2 = 3 / 2 o r 1 .5$

CHXII04:CHEMICAL KINETICS

320535 Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide
$2N O_{(g)} + O_{2} (g) \overset{K}{\to} 2N O_{2(g)}$
If the mechanism of reaction is
$NO + O_{2} \underset{K}{⇌} N_{3} ( fast ) N O_{3} + NO \overset{K_{1}}{\to}$
$N O_{2} + N O_{2} (slow)$
then rate law is

1 Rate

= K^{'} [N O] [O_{2}]

2 Rate

= K^{'} [NO] {[O_{2}]}^{2}

3 Rate

= K^{'} [N O]^{2} [O_{2}]

4 Rate

= K^{'} [N O]^{3} [O_{2}]

Explanation:

From slowest step i.e., rds step,
Rate $= k_{1} [{NO}_{3}] [NO]$
${NO}_{3}$ is an intermediate, need to be replaced with $NO$ and $O_{2}$ by using fast step.
$K = \frac{[N O_{3}]}{[N O] [O_{2}]} \Rightarrow [N O_{3}] = K [N O] [O_{2}]$
Substitute $[{NO}_{3}]$ in eq. (1)
Rate $= k_{1} K [N O] [O_{2}] [N O]$
Rate $= k^{'} [N O]^{2} [O_{2}] [k^{'} = k_{1} K]$

CHXII04:CHEMICAL KINETICS

320536 The plot of $t_{1 / 2} v / s {[R]}_{0}$ for a reaction is a straight-line parallel to $x$ -axis. The unit for the rate constant of this reaction is

1

{molL}^{- 1} s

2

{molL}^{- 1} s^{- 1}

3

{Lmol}^{- 1} s^{- 1}

4

s^{- 1}

Explanation:

Plot of $t_{1 / 2} vs [R]_{0}$ for a reaction is a straight line parallel to $x$ -axis, means $t_{1 / 2} vs [R]_{0}$ is independent of $[R]_{0}$ hence, the reaction is first order.
$∴$ Unit of rate constant $= s^{- 1}$

CHXII04:CHEMICAL KINETICS

320537 The reaction, $X + 2 Y + Z \to N$ occurs by the following mechanism

(i) $X + Y ⇌ M$ very rapid equilibrium)

(ii) $M + Z \to O$ (slow) (iii) $O + Y \to N$ (very fast)

What is the rate law for this reaction?

1

R a t e = k [Z]

2

R a t e = k [X] [Y]^{2} [Z]

3

R a t e = k [N]

4

R a t e = k [X] [Y] [Z]

Explanation:

$K_{C} = \frac{[M]}{[X] [Y]} \Rightarrow [M] = K_{C} [X] [Y]$
$R a t e = k^{'} [M] [Z] = k^{'} K_{C} [X] [Y] [Z] = K [X] [Y] [Z]$

CHXII04:CHEMICAL KINETICS

320538 The reaction, $X_{2} + Y_{2} \to 2 X Y$ occurs by the following mechanism

(i) $X_{2} ⇌ 2 X$ (fast)

(ii) $X + Y_{2} \to X Y + X$ (slow)

(iii) $X + Y \to X Y$ (fast)
What is the order of the reaction?

1 2

2 0

3 -1

4 1.5

Explanation:

From slowest step (rds),
$r = k [X] [Y_{2}]$
From fast step
$K = \frac{{[X]}^{2}}{[X_{2}]}$
$[X] = K^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}}$
$∴ r = k K^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}} [Y_{2}]$
$r = k^{'} {[X_{2}]}^{\frac{1}{2}} [Y_{2}] (k^{'} = k K^{\frac{1}{2}})$
Order of the reaction $= 1 + 1 / 2 = 3 / 2 o r 1 .5$

CHXII04:CHEMICAL KINETICS

320535 Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide
$2N O_{(g)} + O_{2} (g) \overset{K}{\to} 2N O_{2(g)}$
If the mechanism of reaction is
$NO + O_{2} \underset{K}{⇌} N_{3} ( fast ) N O_{3} + NO \overset{K_{1}}{\to}$
$N O_{2} + N O_{2} (slow)$
then rate law is

1 Rate

= K^{'} [N O] [O_{2}]

2 Rate

= K^{'} [NO] {[O_{2}]}^{2}

3 Rate

= K^{'} [N O]^{2} [O_{2}]

4 Rate

= K^{'} [N O]^{3} [O_{2}]

Explanation:

From slowest step i.e., rds step,
Rate $= k_{1} [{NO}_{3}] [NO]$
${NO}_{3}$ is an intermediate, need to be replaced with $NO$ and $O_{2}$ by using fast step.
$K = \frac{[N O_{3}]}{[N O] [O_{2}]} \Rightarrow [N O_{3}] = K [N O] [O_{2}]$
Substitute $[{NO}_{3}]$ in eq. (1)
Rate $= k_{1} K [N O] [O_{2}] [N O]$
Rate $= k^{'} [N O]^{2} [O_{2}] [k^{'} = k_{1} K]$

CHXII04:CHEMICAL KINETICS

320536 The plot of $t_{1 / 2} v / s {[R]}_{0}$ for a reaction is a straight-line parallel to $x$ -axis. The unit for the rate constant of this reaction is

1

{molL}^{- 1} s

2

{molL}^{- 1} s^{- 1}

3

{Lmol}^{- 1} s^{- 1}

4

s^{- 1}

Explanation:

Plot of $t_{1 / 2} vs [R]_{0}$ for a reaction is a straight line parallel to $x$ -axis, means $t_{1 / 2} vs [R]_{0}$ is independent of $[R]_{0}$ hence, the reaction is first order.
$∴$ Unit of rate constant $= s^{- 1}$

CHXII04:CHEMICAL KINETICS

320537 The reaction, $X + 2 Y + Z \to N$ occurs by the following mechanism

(i) $X + Y ⇌ M$ very rapid equilibrium)

(ii) $M + Z \to O$ (slow) (iii) $O + Y \to N$ (very fast)

What is the rate law for this reaction?

1

R a t e = k [Z]

2

R a t e = k [X] [Y]^{2} [Z]

3

R a t e = k [N]

4

R a t e = k [X] [Y] [Z]

Explanation:

$K_{C} = \frac{[M]}{[X] [Y]} \Rightarrow [M] = K_{C} [X] [Y]$
$R a t e = k^{'} [M] [Z] = k^{'} K_{C} [X] [Y] [Z] = K [X] [Y] [Z]$

CHXII04:CHEMICAL KINETICS

320538 The reaction, $X_{2} + Y_{2} \to 2 X Y$ occurs by the following mechanism

(i) $X_{2} ⇌ 2 X$ (fast)

(ii) $X + Y_{2} \to X Y + X$ (slow)

(iii) $X + Y \to X Y$ (fast)
What is the order of the reaction?

1 2

2 0

3 -1

4 1.5

Explanation:

From slowest step (rds),
$r = k [X] [Y_{2}]$
From fast step
$K = \frac{{[X]}^{2}}{[X_{2}]}$
$[X] = K^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}}$
$∴ r = k K^{\frac{1}{2}} {[X_{2}]}^{\frac{1}{2}} [Y_{2}]$
$r = k^{'} {[X_{2}]}^{\frac{1}{2}} [Y_{2}] (k^{'} = k K^{\frac{1}{2}})$
Order of the reaction $= 1 + 1 / 2 = 3 / 2 o r 1 .5$

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