Question

The initial rate of a reaction triples when the concentration of a reactant, ( A ), is doubled. The order of the reaction with respect to ( A ) is _______ (rounded off to 2 decimal places).

Hint:

To determine the reaction order when rates and concentrations are given, use the formula: [ n = frac{ln left( frac{{Rate 2}}{{Rate 1}} right)}{ln left( frac{{Concentration 2}}{{Concentration 1}} right)}. ]

Answer 1

Step 1: Write the rate law for the reaction.

The rate of a reaction can be expressed as:

\[ r = k [A]^n \]

where:

• \( r \) = rate of reaction
• \( k \) = rate constant
• \( [A] \) = concentration of reactant \( A \)
• \( n \) = order of the reaction with respect to \( A \)

Step 2: Analyze the given data.

When the concentration of \( A \) is doubled, the rate of reaction triples. Mathematically:

\[ \frac{r_2}{r_1} = \frac{k [2A]^n}{k [A]^n} \]

Simplify:

\[ \frac{r_2}{r_1} = 2^n \]

Given that \( \frac{r_2}{r_1} = 3 \):

\[ 3 = 2^n \] Step 3: Solve for \( n \).

Take the logarithm on both sides:

\[ \ln(3) = n \ln(2) \] \[ n = \frac{\ln(3)}{\ln(2)} \]

Substituting values:

\[ n = \frac{1.0986}{0.6931} \] \[ n \approx 1.585 \] Step 4: Conclusion.

The order of the reaction with respect to \( A \) is approximately \( 1.585 \), which lies between \( 1.55 \) and \( 1.60 \).