Chemistry Chemical Kinetics

PCl₅(g) -> PCl₃ (g) + Cl₂(g)

In the above first order reaction, the concentration of PCl₅ reduces from initial concentration 50mol L^-1 to mol L^-1 in 120 minutes at 300K. The rate constant for the reaction at 300K is x * 10^-2 min^-1. the value of x is_________.

[Given log5 = 0.6989]

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Contributor-Level 10

$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$ (first order reaction)

$K = \frac{2.303}{120} l o g (\frac{50}{10})$

$∴ K = \frac{2.303 * l o g 5}{120 m i n} = \frac{2.303 * 0.6989}{120}$ = 0.0134 min^-1

Rate constant at 300K = 1.34 * 10^-2min^-1 [the nearest integer = 1.0]

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For the reaction $2 N O_{2} (g) \overset{}{?} N_{2} O_{4} (g), w h e n Δ S = - 176.0 J K^{- 1} a n d Δ H = - 57.8 k J$ mol^-1, the magnitude of $Δ G$ at 298 K for the reaction is______ kJ mol^-1. (Nearest integer)

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Chemistry Chemical Kinetics Chemical Kinetics

Contributor-Level 10

$2 N O_{2} (g) ? N_{2} O_{4} (g)$

$Δ S = - 176 J K^{- 1} = - 0.176 k J K^{- 1}$

$Δ H = - 57.8 K J m o l^{- 1}$

T = 298 K

Using , $Δ G = Δ H - T Δ S$

$= - 57.8 + (298 * 0.176) K J m o l^{- 1}$

$= - 5.3 K J m o l^{- 1}$

Hence, magnitude of $Δ G$ in kJmol^-1 is 5 (nearest integers)

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For the reaction A -> B, the rate constant K(in s^-1) is givne by

$l o g_{10} K = 20.35 - \frac{(2.47 * 1 0^{3})}{T} .$

The energy of activation in KJ mol^-1 is ______. (Nearest integer)

[Given : R = 8.314 J K^-1 mol^-1]

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Contributor-Level 10

log k = 20.35 - $\frac{(2.47 * 1 0^{3})}{T}$

Comparing with,

$l o g k = l o g A - \frac{E a}{2.303 R T}$

$\frac{E a}{2.303 R} = 2.47 * 1 0^{3}$

= 47.29 kJ/mole

Ans. = 47

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For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is________. (Integer answer)

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Raj Pandey

Contributor-Level 9

$Δ H = {(E_{a c t})}_{f} - {(E_{a c t})}_{b}$

= x – (x + y) = -y

= -45 KJ/mol

Ans. = 45

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The first order rate constant for the decomposition of CaCO₃ at 700 K is 6.36 * 10^-3 s^-1 and activation energy is 209 kJ mol^-1. Its rate constant (in s^-1) at 600K is x * 10^-6. The value of x is_____. (nearest integer).

$[G i v e n R = 8.31 J K^{- 1} m o l^{- 1}; l o g 6.36 * 1 0^{- 3} = - 2.19, 1 0^{- 4.79} = 1.62 * 1 0^{- 5}]$

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Contributor-Level 10

$K_{700} = 6.36 * 1 0^{- 3} s^{- 1}$

$K_{600} = x * 1 0^{- 6} s^{- 1}$

E_a = 209 KJ/mol

$l o g (\frac{K_{2}}{K_{1}}) = \frac{- E a}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

$l o g K_{600} = - 4.79$

$K_{600} = 1 0^{- 4.79} = 1.62 * 1 0^{- 5}$

= 16.2 * 10^-6

x = 16 (the nearest integer)

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At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340s. When the pressure was 27.8 kPa., the half life was found to be 170s. The order of the reaction is ________. [integer answer]

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Raj Pandey

Contributor-Level 9

t1/2 = 340 secP0 = 55.5 kPa

t1/2 = 170 sec P0 = 27.8 kPa

$∴ t_{1 / 2} \propto {(P_{0})}^{1 - n}$

$∴ 1 - n = 1 n = 0$

$∴$ zero order reaction

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The following data was obtained for chemical reaction given below at 975 K.

2NO_(g) + 2H_2(g) -> N_2(g) + 2H₂O_(g)

[NO] [H₂] &

...more

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Contributor-Level 10

If Rate = $K {[N O]}^{a} {[H_{2}]}^{b}$

Case I => $7 * 1 0^{- 9} = K {(8 * 1 0^{- 5})}^{a} {(8 * 1 0^{- 5})}^{b}$

Case II => $2.1 * 1 0^{- 8} = k {(24 * 1 0^{- 5})}^{a} {(8 * 1 0^{- 5})}^{b}$

Therefore, $\frac{7 ? * 1 0^{- 1}}{2.1} = {(\frac{1}{3})}^{a}$

or ${(\frac{1}{3})}^{1} = {(\frac{1}{3})}^{a} ∴ a = 1$

Ans. = 1

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5 months ago

The reaction rate for the reaction

${[P t C l_{4}]}^{2 -} + H_{2} O \overset{}{?} {[P t (H_{2} O) C l_{3}]}^{-} + C l^{-}$

was measured as a function of concentrations of different species. It was observed that

$\frac{- d {[[P t C l_{4}]]}^{2 -}}{d t} = 4.8 * 1 0^{- 5} [{[P t C l_{4}]}^{2 -}] - 2.4 * 1 0^{- 3} [{[P t (H_{2} O) C l_{3}]}^{-}] [C l^{-}]$

where square bracket are used to denote molar concentrations. The equilibrium constant K_C= ________. (Nearest integer)

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Contributor-Level 10

${[P t C l_{4}]}^{2 -} + H_{2} O ? {[P t (H_{2} O) C l_{3}]}^{-} + C l^{-}$

At equilibrium $\frac{d x}{d t} = 0,$ hence

$0 = \frac{- d [{[P t C l_{4}]}^{2 -}]}{d t} = 4.8 * 1 0^{- 5} [{[P t C l_{4}]}^{2 -}] - 2.4 * 1 0^{- 3} [{[P t (H_{2} O) C l_{3}]}^{-}] [C l^{-}]$

$K_{c} = \frac{K_{f}}{K_{b}} = \frac{4.8 * 1 0^{- 5}}{2.4 * 1 0^{- 3}}$

= 2 * 10^-2

= 0.02

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Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Heavy water is used for the study of reaction mechanism.

Reason (R): The rate of reaction for the cleavage of O-H bond is slower than that of O-D bond.

Choose the most appropriate answer from the options given below:

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Chemistry Chemical Kinetics Chemical Kinetics

Contributor-Level 10

Heavy water (D₂O) is used to study reaction mechanism. O-H bond energy is less than O-D, hence it will produce fast reaction than O-D. Here assertion is true but Reason is false.

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