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Bromate ions react with bromide ions and protons to form bromine and water. Broj + Br + _ H → _ Br2 + H2O a. Balance the reac

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Answer #1

4.

a)

BrO3- + 5 Br- + 6 H +\to 3 Br2 + 3 H2O

b)

rate law is

rate = K × [BrO3- ]x [Br-]y [H+]z

from the given values

5.4×10-3 = K (0.20)x(0.10)y (0.15)z ..[1]

2.7×10-3 = K (0.10)x(0.10)y(0.15)z ...[2]

1.08×10-2 = K(0.20)x (0.40)y (0.15)z ...[3].

1.35×10-3 = K(0.20)x(0.10)y(0.075)z ...[4].

Eq.1 ÷ Eq.2

(5.4×10-3/2.7×10-3) = ( 0.20/0.10)x

or, 2 = 2x

or, x =1

Eq.3÷Eq.2

(1.08×10-2​​​​​​/ 5.4×10-3) = (0.4/0.1)y

or, (2) = 4 y

or, y = 0.5

Eq.1 ÷Eq.4.

(5.4×10-3/1.35×10-3) = (0.15/0.075)z

or, 4 = (2)z

or, z = 2.

Hence rate law is

Rate =K ×[BrO3- ] [ Br-]0.5 [H+ ]2.

overall order = (1+0.5+2) = 3.5

C)

using Eq.1

5.4×10-3 = K × (0.20) (0.10)0.5(0.15)2

or, 5.4×10-3 = K × 0.0014

or, K = 3.86 mol(-2.5) L(2.5) s-1.

d.

From the balanced reaction

rate = - \frac{1}{5} di Br- = + \frac{1}{3}   dBral (1)

Or,

given, - di Br- = 2.5 M/s

So, from Eq.1

Rate of appearance of Br2 = + dBral = - \frac{3}{5} di Br-

= \frac{3}{5} × 2.5 M/s

= 1.5 M/s.

e)

Rate = K × [BrO3- ] [Br ]0.5[H+] 2

Let initial [BrO3] = a , [Br-] = b, [H+] = c

Rate1 = K × (a)​​​​​​ (b )0.5 (c )2

When, [BrO3-] is doubled and [H+] is reduced by (1/2)

Rate2 = K × (2×a)× (b )0.5 (c/2)2

Rate1/Rate2

= 2 × (\frac{1}{4})

= \frac{1}{2}

Hence, the rate will be reduced to half.

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