Answer :
To determine the rate law for the reaction [tex]\( A + 2B \rightarrow D \)[/tex], we need to examine the experimental observations provided:
1. The rate of formation of [tex]\( D \)[/tex] is independent of the concentration of [tex]\( B \)[/tex].
2. The rate of formation of [tex]\( D \)[/tex] quadruples when the concentration of [tex]\( A \)[/tex] is doubled.
Given these observations, let's develop the detailed solution step-by-step:
### Step 1: Independent of B
First, we interpret the observation that the rate of formation of [tex]\( D \)[/tex] is independent of the concentration of [tex]\( B \)[/tex]. This implies that changes in the concentration of [tex]\( B \)[/tex] do not affect the rate of the reaction. In terms of the rate law, this means that the reaction order with respect to [tex]\( B \)[/tex] is zero.
For a reactant [tex]\( B \)[/tex] with zero-order dependence, the rate law will not include the concentration of [tex]\( B \)[/tex]:
[tex]\[ \text{rate} = k \times [A]^n \times [B]^0 \][/tex]
Since anything raised to the power of zero is 1, this simplifies to:
[tex]\[ \text{rate} = k \times [A]^n \][/tex]
### Step 2: Dependence on A
Next, we consider the observation that the rate quadruples (i.e., increases by a factor of 4) when the concentration of [tex]\( A \)[/tex] is doubled. Mathematically, we set up the relationship between the concentration of [tex]\( A \)[/tex] and the rate:
- Let the initial concentration of [tex]\( A \)[/tex] be [tex]\([A]\)[/tex].
- Let the rate law be given by [tex]\( \text{rate} = k \times [A]^n \)[/tex].
If the concentration of [tex]\( A \)[/tex] is doubled, [tex]\([A]\)[/tex] becomes [tex]\( 2[A] \)[/tex].
According to the observation, the rate of formation of [tex]\( D \)[/tex] quadruples when the concentration of [tex]\( A \)[/tex] is doubled:
[tex]\[ 4 \times \text{rate} = k \times (2[A])^n \][/tex]
We know that originally:
[tex]\[ \text{rate} = k \times [A]^n \][/tex]
### Step 3: Solving for n
Now, equate the quadrupled rate to the original rate to solve for the exponent [tex]\( n \)[/tex]:
[tex]\[ 4 \times k \times [A]^n = k \times (2[A])^n \][/tex]
[tex]\[ 4 \times [A]^n = (2[A])^n \][/tex]
Simplifying the right-hand side:
[tex]\[ (2[A])^n = 2^n \times [A]^n \][/tex]
Thus, our equation becomes:
[tex]\[ 4 \times [A]^n = 2^n \times [A]^n \][/tex]
We can divide both sides by [tex]\( [A]^n \)[/tex]:
[tex]\[ 4 = 2^n \][/tex]
To find [tex]\( n \)[/tex]:
[tex]\[ 4 = 2^n \][/tex]
We recognize that [tex]\( 4 = 2^2 \)[/tex]. Therefore:
[tex]\[ 2^2 = 2^n \][/tex]
Thus:
[tex]\[ n = 2 \][/tex]
### Step 4: Writing the rate law
Now that we have the values of [tex]\( n \)[/tex] and the order with respect to [tex]\( B \)[/tex] (which is zero), we can write the final rate law for the reaction:
[tex]\[ \text{rate} = k \times [A]^2 \][/tex]
### Conclusion
Therefore, the rate law for the reaction [tex]\( A + 2B \rightarrow D \)[/tex] is:
[tex]\[ \text{rate} = k \times [A]^2 \][/tex]
This is consistent with the experimental observations provided.
1. The rate of formation of [tex]\( D \)[/tex] is independent of the concentration of [tex]\( B \)[/tex].
2. The rate of formation of [tex]\( D \)[/tex] quadruples when the concentration of [tex]\( A \)[/tex] is doubled.
Given these observations, let's develop the detailed solution step-by-step:
### Step 1: Independent of B
First, we interpret the observation that the rate of formation of [tex]\( D \)[/tex] is independent of the concentration of [tex]\( B \)[/tex]. This implies that changes in the concentration of [tex]\( B \)[/tex] do not affect the rate of the reaction. In terms of the rate law, this means that the reaction order with respect to [tex]\( B \)[/tex] is zero.
For a reactant [tex]\( B \)[/tex] with zero-order dependence, the rate law will not include the concentration of [tex]\( B \)[/tex]:
[tex]\[ \text{rate} = k \times [A]^n \times [B]^0 \][/tex]
Since anything raised to the power of zero is 1, this simplifies to:
[tex]\[ \text{rate} = k \times [A]^n \][/tex]
### Step 2: Dependence on A
Next, we consider the observation that the rate quadruples (i.e., increases by a factor of 4) when the concentration of [tex]\( A \)[/tex] is doubled. Mathematically, we set up the relationship between the concentration of [tex]\( A \)[/tex] and the rate:
- Let the initial concentration of [tex]\( A \)[/tex] be [tex]\([A]\)[/tex].
- Let the rate law be given by [tex]\( \text{rate} = k \times [A]^n \)[/tex].
If the concentration of [tex]\( A \)[/tex] is doubled, [tex]\([A]\)[/tex] becomes [tex]\( 2[A] \)[/tex].
According to the observation, the rate of formation of [tex]\( D \)[/tex] quadruples when the concentration of [tex]\( A \)[/tex] is doubled:
[tex]\[ 4 \times \text{rate} = k \times (2[A])^n \][/tex]
We know that originally:
[tex]\[ \text{rate} = k \times [A]^n \][/tex]
### Step 3: Solving for n
Now, equate the quadrupled rate to the original rate to solve for the exponent [tex]\( n \)[/tex]:
[tex]\[ 4 \times k \times [A]^n = k \times (2[A])^n \][/tex]
[tex]\[ 4 \times [A]^n = (2[A])^n \][/tex]
Simplifying the right-hand side:
[tex]\[ (2[A])^n = 2^n \times [A]^n \][/tex]
Thus, our equation becomes:
[tex]\[ 4 \times [A]^n = 2^n \times [A]^n \][/tex]
We can divide both sides by [tex]\( [A]^n \)[/tex]:
[tex]\[ 4 = 2^n \][/tex]
To find [tex]\( n \)[/tex]:
[tex]\[ 4 = 2^n \][/tex]
We recognize that [tex]\( 4 = 2^2 \)[/tex]. Therefore:
[tex]\[ 2^2 = 2^n \][/tex]
Thus:
[tex]\[ n = 2 \][/tex]
### Step 4: Writing the rate law
Now that we have the values of [tex]\( n \)[/tex] and the order with respect to [tex]\( B \)[/tex] (which is zero), we can write the final rate law for the reaction:
[tex]\[ \text{rate} = k \times [A]^2 \][/tex]
### Conclusion
Therefore, the rate law for the reaction [tex]\( A + 2B \rightarrow D \)[/tex] is:
[tex]\[ \text{rate} = k \times [A]^2 \][/tex]
This is consistent with the experimental observations provided.
