8H- - 4Cl- - MnO4- 2Cl2 - Mn2- - 4H2OTrial -Cl- -MnO4-Rate M-s 1 0-0104 0-00800 2-25- 10-13 2 0-0312 0-00800 2-03- 10-12 3 0-0312 0-00400 1-02- 10-12Based on the above information- determine the overall rate law for the reaction given that the reaction is third order with respect to the hydrogen ion-

The correct option is A Rate = k[Cl^ ]^2[MnO_4^ ][H^+]^3 Rate #160; =K[ Cl ^ #160;] #160; ^ m [ MnO _ 4 ^ #160;] #160; ^ n [ H ^ + ...

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Byju's Answer

Standard XII

Chemistry

Rate of Reaction

8H+ + 4Cl- + ...

Question

$8 H^{+} + 4 C l^{-} + M n O_{4}^{-} \to 2 C l_{2} + M n^{2 +} + 4 H_{2} O$
Trial $[C l^{-}]$ $[M n O_{4}^{-}]$ Rate (M/s)
$1$ $0.0104$ $0.00800$ $2.25 \times 10^{- 13}$
$2$ $0.0312$ $0.00800$ $2.03 \times 10^{- 12}$
$3$ $0.0312$ $0.00400$ $1.02 \times 10^{- 12}$
Based on the above information, determine the overall rate law for the reaction given that the reaction is third order with respect to the hydrogen ion.

R a t e = k [C l^{-}]^{2} [M n O_{4}^{-}] [H^{+}]^{3}

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R a t e = k [C l^{-}]^{2} [M n O_{4}^{-}]^{3} [H^{+}]^{3}

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R a t e = k [C l^{-}] [M n O_{4}^{-}]^{3} [H^{+}]^{3}

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R a t e = k [C l^{-}]^{3} [M n O_{4}^{-}] [H^{+}]^{3}

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Solution

The correct option is A $R a t e = k [C l^{-}]^{2} [M n O_{4}^{-}] [H^{+}]^{3}$
Rate $= K {[{C l}^{-}]}^{m} {[{M n O}_{4}^{-}]}^{n} {[H^{+}]}^{3}$
${(\frac{0.0312}{0.0312})}^{m} {(\frac{0.00800}{0.00400})}^{n} = \frac{2.03 \times 10^{- 12}}{1.02 \times 10^{- 12}}$
$2^{n} = 2$
$n = 1$
${(\frac{0.0104}{0.0312})}^{m} {(\frac{0.00800}{0.00800})}^{n} = \frac{2.25 \times 10^{- 13}}{2.03 \times 10^{- 12}}$
${(0.33)}^{m} = 1.10 \times 10^{- 1}$
$m = 2$
Rate $= {K [{C l}^{-}]}^{2} [{M n O}_{4}^{-}] {[H^{+}]}^{3}$

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Similar questions

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

(i) 3 NO(g) → N₂O(g) Rate = k[NO]²

(ii) H₂O₂(aq) + 3 I⁻(aq) + 2 H⁺ → 2 H₂O (l) + Rate = k[H₂O₂][I⁻]

(iii) CH₃CHO(g) → CH₄(g) + CO(g) Rate = k [CH₃CHO]^3/2

(iv) C₂H₅Cl(g) → C₂H₄(g) + HCl(g) Rate = k [C₂H₅Cl]

Q. For the redox reaction,

M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} \to M n^{2} + C O_{2} + H_{2} O

The correct coefficients of the reactants for the balanced equation are:

Q. The data below are for the reaction of

N O

and

{C l}_{2}

to form

N O C l

295

K

Expt. No.	$[{C l}_{2}]$ $(m o l L^{- 1})$	$[N O]$ $(m o l L^{- 1})$	Initial rate $(m o l L^{- 1} s^{- 1})$
$1.$	$0.05$	$0.05$	$1.0 \times 10^{- 3}$
$2.$	$0.15$	$0.05$	$3.0 \times 10^{- 3}$
$3.$	$0.05$	$0.15$	$9.0 \times 10^{- 3}$

(a) What is the order with respect to

N O

and

{C l}_{2}

in the reaction?
(b) Write the rate expression.
(c) Calculate the rate constant.
(d) Determine the reaction rate when concentration of

{C l}_{2}

and

N O

are

0.2

M

and

0.4

M

respectively.

C {C l}_{3} C H O + N O ⟶ C H {C l}_{3} + N O + C O

The rate of the reaction above is equal to rate

= k [C {C l}_{3} C H O] [N O]

If the concentration is expressed in mol L

^{- 1}

, find the unit of

k

Q. Consider the reaction:

{C l}_{2} (a q) + H_{2} S (a q) \to S (s) + 2 H^{+} (a q) + 2 {C l}^{-} (a q)

The rate equation for this reaction is
rate

= k [{C l}_{2}] [H_{2} S]

Which of these mechanisms is/are consistent with this rate equation?
A.

{C l}_{2} + H_{2} S \to H^{+} + {C l}^{-} + {C l}^{+} + {H S}^{-}

(slow)

{C l}^{+} + {H S}^{-} \to H^{+} + {C l}^{-} + S

(fast)

H_{2} S ⇌ H^{+} + {H S}^{-}

(fast equilibrium)

{C l}_{2} + {H S}^{-} \to 2 {C l}^{-} + H^{+} + S

(Slow)

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Trial	$[C l^{-}]$	$[M n O_{4}^{-}]$	Rate (M/s)
$1$	$0.0104$	$0.00800$	$2.25 \times 10^{- 13}$
$2$	$0.0312$	$0.00800$	$2.03 \times 10^{- 12}$
$3$	$0.0312$	$0.00400$	$1.02 \times 10^{- 12}$

8H++4Cl−+MnO4−→2Cl2+Mn2++4H2OTrial[Cl−][MnO4−]Rate (M/s)10.01040.008002.25×10−1320.03120.008002.03×10−1230.03120.004001.02×10−12Based on the above information, determine the overall rate law for the reaction given that the reaction is third order with respect to the hydrogen ion.

The correct option is A Rate=k[Cl−]2[MnO4−][H+]3Rate =K[Cl−]m[MnO−4]n[H+]3(0.03120.0312)m(0.008000.00400)n=2.03×10−121.02×10−12 2n=2 n=1(0.01040.0312)m(0.008000.00800)n=2.25×10−132.03×10−12 (0.33)m=1.10×10−1 m=2Rate =K[Cl−]2[MnO−4][H+]3