Question

1. the following steps are involved in a reaction:

(fast) 1. ( cl_{2(g)}\to2cl_{(g)} )
(slow) 2. ( chcl_{3(g)}+cl_{(g)}\to hcl_{(g)}+ccl_{3(g)} )
(fast) 3. ( ccl_{3(g)}+cl_{(g)}\to ccl_{4(g)} )
a) state which equation is the rate determining step (r.d.s.).
b) write the overall reaction equation for this reaction mechanism.
c) write the rate law for this reaction mechanism.

Explanation:

Step1: Identify Rate Determining Step (R.D.S.)

In a reaction mechanism, the slowest step is the rate - determining step. Given the steps:

1. \(Cl_{2(g)}\to2Cl_{(g)}\) (Fast)
2. \(CHCl_{3(g)}+Cl_{(g)}\to HCl_{(g)}+CCl_{3(g)}\) (Slow)
3. \(CCl_{3(g)}+Cl_{(g)}\to CCl_{4(g)}\) (Fast)

Since step 2 is the slowest, it is the rate - determining step.

Step2: Write the overall reaction equation

Add the three steps together:

• Step 1: \(Cl_{2(g)}\to2Cl_{(g)}\)
• Step 2: \(CHCl_{3(g)}+Cl_{(g)}\to HCl_{(g)}+CCl_{3(g)}\)
• Step 3: \(CCl_{3(g)}+Cl_{(g)}\to CCl_{4(g)}\)

Cancel out the intermediate species (\(Cl\) and \(CCl_{3}\)):
\(Cl_{2(g)}+CHCl_{3(g)}\to HCl_{(g)}+CCl_{4(g)}\)

Step3: Write the rate law

The rate law is based on the rate - determining step (step 2). The rate law for a reaction \(A + B\to products\) is \(rate = k[A][B]\) (for elementary steps).
For step 2: \(CHCl_{3(g)}+Cl_{(g)}\to HCl_{(g)}+CCl_{3(g)}\), the rate law is \(rate = k[CHCl_{3}][Cl]\)

But \(Cl\) is an intermediate. From step 1 (\(Cl_{2(g)}
ightleftharpoons2Cl_{(g)}\)), using the equilibrium constant \(K=\frac{[Cl]^{2}}{[Cl_{2}]}\), we can express \([Cl]=K^{\frac{1}{2}}[Cl_{2}]^{\frac{1}{2}}\)

Substitute \([Cl]\) into the rate law:
\(rate = k'[CHCl_{3}][Cl_{2}]^{\frac{1}{2}}\) where \(k' = kK^{\frac{1}{2}}\)

Answer:

a) The rate - determining step is \(CHCl_{3(g)}+Cl_{(g)}\to HCl_{(g)}+CCl_{3(g)}\)
b) The overall reaction equation is \(Cl_{2(g)}+CHCl_{3(g)}\to HCl_{(g)}+CCl_{4(g)}\)
c) The rate law is \(rate = k[CHCl_{3}][Cl_{2}]^{\frac{1}{2}}\)