Question

Can you please explain how to obtain answers to all parts of this question? Please provide step by step instructions. I'm very confused on how to predict this information. Predict the rate law for the reaction NO + Br2 --> NOBr2 if : a. the rate doubles when [NO] is doubled and [Br2] remains constant. b. the rate doubles when [Br2] is doubled and [NO] remains constant. c. The rate increase by 1.56 times when [NO] is increased 1.25 times and [Br2] remains constant. d. The rate is halved when [NO] is doubled and [Br2] remains constant.

Answer 1

The rate law for the given reaction NO + Br₂ --> NOBr₂ can be given as-

Rate = k * [NO]^x * [Br₂]^y

where

• k = rate constant of the reaction
• x = order of reaction with respect to reactant NO
• y = order of reaction with respect to reactant Br₂

Now for-

if : a.

the rate doubles when [NO] is doubled and [Br2] remains constant

That means lets take two conditions-

Condition 1 : [NO] = A, [Br2] = B, and rate = R1

Condition 2 : [NO] = 2A, [Br2] = B, and rate = 2R1 (since rate doubles)

Now if we write the rate law for condition 1, then it will be-

Rate1 = k1 * [NO]^x * [Br₂]^y

R1 = k1 * [A]^x * [B]^y

Similarly

if we write the rate law for condition 2, then it will be-

Rate1 = k1 * [NO]^x * [Br₂]^y

2R1 = k2 * [2A]^x * [B]^y

Now dividing the two tares i.e

Rate1/Rate2 = {k1 * [A]^x * [B]^y} / {k2 * [2A]^x * [B]^y}

R1/2R1 = (k1/k2) * ([A]^x / [2A]^x) * ([B]^y / [B]^y)

1/2 = (A/2A)^x since the rate constant of a reaction has a fixed value, we can say k1 = k2

1/2 = (1/2)^x

(1/2)¹ = (1/2)^x

x = 1

That means here rate law will be-

Rate = k * [NO]^x * [Br₂]^y

Rate = k * [NO]¹ * 1 here we can omit [B] as its concentration does not change at all.

Rate = k * [NO]

b- Similarly  

the rate doubles when [Br2] is doubled and [NO] remains constant

That means lets take two conditions-

Condition 1 : [Br2] = B, [NO] = A, and rate = R1

Condition 2 : [Br2] = 2B, [NO] = A, and rate = 2R1 (since rate doubles)

Now if we write the rate law for condition 1, then it will be-

Rate1 = k1 * [A]^x * [B]^y

R1 = k1 * [A]^x * [2B]^y

Similarly

if we write the rate law for condition 2, then it will be-

Rate1 = k1 * [NO]^x * [Br₂]^y

2R1 = k2 * [A]^x * [2B]^y

Now dividing the two rates i.e

Rate1/Rate2 = {k1 * [A]^x * [B]^y} / {k2 * [A]^x * [2B]^y}

R1/2R1 = (k1/k2) * ([A]^x / [A]^x) * ([B]^y / [2B]^y)

1/2 = (B/2B)^y since the rate constant of a reaction has a fixed value, we can say k1 = k2

1/2 = (1/2)^y

(1/2)¹ = (1/2)^y

y = 1

That means here rate law will be-

Rate = k * [NO]^x * [Br₂]^y

Rate = k * [Br2]¹ * 1 here we can omit [A] as its concentration does not change at all.

Rate = k * [Br2]

c-For

The rate increase by 1.56 times when [NO] is increased 1.25 times and [Br2] remains constant

That means lets take two conditions-

Condition 1 : [NO] = A, [Br2] = B, and rate = R1

Condition 2 : [NO] = 1.25A, [Br2] = B, and rate = 1.56R1 (since rate doubles)

Now if we write the rate law for condition 1, then it will be-

Rate1 = k1 * [NO]^x * [Br₂]^y

R1 = k1 * [A]^x * [B]^y

Similarly

if we write the rate law for condition 2, then it will be-

Rate1 = k1 * [NO]^x * [Br₂]^y

1.56R1 = k2 * [1.25A]^x * [B]^y

Now dividing the two tares i.e

Rate1/Rate2 = {k1 * [A]^x * [B]^y} / {k2 * [2A]^x * [B]^y}

R1/1.56R1 = (k1/k2) * ([A]^x / [1.25A]^x) * ([B]^y / [B]^y)

1/1.56 = (A/1.25A)^x since the rate constant of a reaction has a fixed value, we can say k1 = k2

1/1.56 = (1/1.25)^x

(0.64)¹ = (0.8)^x

(0.8)² = (0.8)^x

x = 2

That means here rate law will be-

Rate = k * [NO]^x * [Br₂]^y

Rate = k * [NO]² * 1 here we can omit [B] as its concentration does not change at all.

Rate = k * [NO]²

d-

The rate is halved when [NO] is doubled and [Br2] remains constant.

That means lets take two conditions-

Condition 1 : [NO] = A, [Br2] = B, and rate = R1

Condition 2 : [NO] = 2A, [Br2] = B, and rate = 1/2R1 (since rate doubles)

Now if we write the rate law for condition 1, then it will be-

Rate1 = k1 * [NO]^x * [Br₂]^y

R1 = k1 * [A]^x * [B]^y

Similarly

if we write the rate law for condition 2, then it will be-

Rate1 = k1 * [NO]^x * [Br₂]^y

1/2R1 = k2 * [2A]^x * [B]^y

Now dividing the two tares i.e

Rate1/Rate2 = {k1 * [A]^x * [B]^y} / {k2 * [2A]^x * [B]^y}

R1/1/2R1 = (k1/k2) * ([A]^x / [2A]^x) * ([B]^y / [B]^y)

2 = (A/2A)^x since the rate constant of a reaction has a fixed value, we can say k1 = k2

2 = (1/2)^x

(1/2)^-1 = (1/2)^x

x = -1

That means here rate law will be-

Rate = k * [NO]^x * [Br₂]^y

Rate = k * [NO]^-1 * 1 here we can omit [B] as its concentration does not change at all.

Rate = k / [NO]