Kinetics of Chain Reaction Between Hydrogen and Bromine

Kinetics of Chain Reaction Between Hydrogen and Bromine

A thermal reaction between hydrogen and bromine could be best comprehend by studying the kinetics of chain reaction that occurs between them.

Step 1: R1 = k1 [Br2]

Step 2: R2 = k2 [H2] [Br]

Step 3: R3 = k3 [H] [Br2]

Step 4: R4 = k4 [H] [HBr]

Step 5: R5 = k5 [Br]²

In above steps, rate of reaction will be equal to the net rate of formation and decomposition of hydrogen bromide.

Rate of reaction = R2 + R3 - R4 eq.21.1

or k2 [H2] [Br] + k3 [H] [Br2] - k4 [H] [HBr]

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Similarly, we can find net rate of reaction formation of bromine and hydrogen separately.

For Bromine, rate of reaction will be equal to the net rate of formation and decomposition of bromine.

R1 -R2 + R3 + R4 - R5

or R1 + R3 + R4 = R2 + R5

Putting values of R1-5, we will get;

k1 [Br2] + k3 [H] [Br2] + k4 [H] [HBr] = k2 [H2] [Br] + k5 [Br]²eq. 21.2

For Hydrogen, rate of reaction will be equal to the net rate of formation and decomposition of hydrogen.

R2 - R3 - R4

or R2 = R3 + R4

Putting values we got,

k2 [H2] [Br] = k3 [H] [Br2] + k4 [H] [HBr] eq.21.3

Adding eq.21.2 and 21.3 (LHS with LHS and RHS with RHS) we got,

k1[Br2] +k3[H][Br2] + k4 [H][HBr] + k2[H2][Br] = k2[H2][Br] + k5[Br]²+ k3[H][Br2] + k4 [H][HBr]

Simplifying,

k1[Br2] = k5[Br]²

Taking square root on both sides we got concentration of bromine as,

eq.21.4

To find concentration of hydrogen, we will use eq.21.3.

k2 [H2] [Br] = k3 [H] [Br2] + k4 [H] [HBr]

Taking [H] common in the above equation we got,

k2 [H2] [Br] = [H] (k3[Br2] + k4[HBr])

eq.21.5

Eq. 21.4 and 21.5 represents the concentration of bromine and hydrogen respectively, Further we can put value of Br in eq.21.5 and then put values of both hydrogen and bromine in eq.21.1 to find the concentration of HBr. Now firstly, putting value of bromine in eq.21.5 we got,

Now following eq.21.1, we can take [H] as common the equation will becomes

k2 [H2] [Br] + (k3 [Br2] - k4[HBr]) [H]

Now we can put values of [H] and [Br] in above equation.

Taking common we got,

Simplifying by taking LCM.

Dividing numerator and denominator by k3[Br2]

eq. 21.6

The eq.21.6 is representing a thermal reaction between hydrogen and bromine. Here HBr is in excess and Br2 is negligible. In the step 3 and 4, there is a competition between Bromine [Br2] and Hydrogen bromide [HBr] for [H]. At the start of reaction, there is no product [HBr], so step 4 can be neglected. Thus eq. 21.6 becomes.