Chemistry Chemical Kinetics

2NO + 2H₂ ® N₂ + 2H₂O

The above reaction has been studied at 800°C. The related data are given in the table below.

The order to the reaction with respect to NO is_______________.

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Contributor-Level 10

On decreasing pressure of NO by a factor of '2' the rate of reaction decreases by a factor of '4'

$∴$ Order of reaction w.r.t 'NO' = 2

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The rate constant for a first order reaction is given by the following equation Ink = 33.24 - $\frac{2.0 * 1 0^{4} K}{T}$

The activation energy for the reaction is given by ___________kJ mol^-¹ (In nearest integer) (Given: R = 8.3JK^-¹mol^-¹)

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According to Arrhenius equation ;

$I n k = l n A - \frac{E a}{R T}$

Here lnk = 33.24 $- \frac{2.0 * 1 0^{4}}{T}$

$∴ \frac{E a}{R} = 2.0 * 1 0^{4}$

$a n d E_{a} = 2.0 * 1 0^{4} * R = 2.0 * 1 0^{4} * 8.3 = 16.6 * 1 0^{4} J = 166 k J$

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For the reaction

$2 H_{2} (g) + 2 N O (g) ? N_{2} (g) + 2 H_{2} O$ , the observed rate expression is, rate = $k_{f} [N O]^{2} [H_{2}]$ . The rate expression for the reverse reaction is:

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Chemistry Chemical Kinetics Chemical Kinetics

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Al equilibrium

$K_{f} [H_{2}] [N O]^{2} = \frac{K_{b} [N_{2}] {[H_{2} O]}^{+}}{ [H_{2}]}$

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[A] -> [B]

Reactant Product

If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x * 10^-⁶ s^-¹. The value of x is___________. (Nearest Integer)

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Contributor-Level 10

$K = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{70 * 60} = \frac{6930}{7 * 6} * 1 0^{- 6}$

= 165 * 10^-6 s^-1

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2NO + 2H₂ → N₂ + 2H₂O

The above reaction has been studied at 800°C. The related data are given in the table below.

Reaction serial number	Initial Pressure of H₂ / kPa	Initial Pressure of NO / kPa	Initial rate (-dP/dT)/(kPa/s)
1	65.6	40.0	0.135
2	65.6	20.1	0.033
3.	38.6	65.6	0.214
4.	19.2	65.6	0.106

The order to the reaction with respect to NO is_______________.

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Payal Gupta

Contributor-Level 10

On decreasing pressure of NO by a factor of '2' the rate of reaction decreases by a factor of '4'

$∴$ Order of reaction w.r.t 'NO' = 2

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New answer posted

5 months ago

Chemistry Chemical Kinetics Chemical Kinetics

For the decomposition of azomethane,

CH₃N₂CH₃(g) → CH₃CH₃(g) + N₂(g), a first order reaction, the variation in partial pressure with time at 600 K is given as.

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Payal Gupta

Contributor-Level 10

For first order reaction,

$l n (\frac{P}{P_{0}}) = - k . t$

$K = 3.465 * 1 0^{4}$

$t_{1 / 2} = \frac{0.693}{3.465 * 1 0^{4}} = 2 * 1 0^{- 5}$ sec

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5 months ago

The half life for the decomposition of gaseous compound A is 240s when the gaseous pressure was 500 Torr initially, When the pressure was 250 Torr, the half life was found to be 4.0 min. The order of the reaction is _________. (Nearest integer)

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Chemistry Chemical Kinetics Chemical Kinetics

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$\frac{{(t_{1 / 2})}_{l}}{{(t_{1 / 2})}_{I I}} = {(\frac{P_{I}}{P_{I I}})}^{1 - n} \Rightarrow n = 1$

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6 months ago

For the given first order reaction

A -> B

The half life of the reaction is 0.3010 min. The ration of the initial concentration of reactant to the concentration of reactant at time 2.0 min will be equal to________________. (Nearest integer)

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Contributor-Level 10

t1/2 = 0.301 min

t = 2 min

$K = \frac{2.303}{t} l o g (\frac{C o}{C t})$

$\frac{0.693}{0.301} = \frac{2.303}{2} l o g (\frac{C o}{C t})$

$\frac{2.303 * 0.301}{0.301} = \frac{2.303}{2} l o g (\frac{C o}{C t})$

$∴ 2 = l o g (\frac{C o}{C t})$

$\frac{C o}{C t} = 1 0^{2} = 100$

Ans. 100

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6 months ago

The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700 – 1000 K. The data has been analysed by plotting ln k vs $\frac{1 0^{3}}{T}$ graph. The value of activation energy for the reaction is ___________ kJ mol^-¹.

(Nearest integer)

(Given : R = 8.31 J K^-¹ mol^-¹)

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Contributor-Level 10

$k = A . e^{- (\frac{E a}{R T})}$

$l n k = l n A - \frac{E a}{R T}$

$\Rightarrow l n k = l n A + (- \frac{E a}{1000 R}) . \frac{1000}{T}$

Slope = $\frac{- E a}{1000 R} = - 18.5$

=> Ea = 18.5 * 1000 * 8.31 = 153.735 * 10³ J = 154 KJ

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6 months ago

For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is

(Given : In 10 = 2.303 and log 2 = 0.3010)

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