The rate law for a reaction between substance \(P\) and \(Q\) is given by
\(\text{Rate = }k[P]^x[Q]^y\)
If the concentration of \(P\) is doubled and the concentration of \(Q\) is halved, the ratio of the new rate to the earlier rate of reaction will be:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. \(2^{x - y}\).
The given rate law for the reaction between substances \(P\) and \(Q\) is:
\(\text{Rate} = k[P]^x[Q]^y\)
The initial rate of reaction is given by:
\(\text{Rate}_\text{initial} = k[P]^x[Q]^y\)
According to the problem:
The concentration of \(P\) is doubled: \(P \to 2[P]\).
The concentration of \(Q\) is halved: \(Q \to \frac{1}{2}[Q]\).
Substituting these changes into the rate law:
\(\text{Rate}_\text{new} = k(2[P])^x\left(\frac{1}{2}[Q]\right)^y\)
This simplifies to:
\(\text{Rate}_\text{new} = k \cdot 2^x [P]^x \cdot \left(\frac{1}{2}\right)^y [Q]^y\)
\(\text{Rate}_\text{new} = k \cdot 2^x \cdot 2^{-y} [P]^x [Q]^y\)
\(\text{Rate}_\text{new} = k \cdot 2^{x - y} [P]^x [Q]^y\)
The ratio of the new rate to the initial rate is:
\(\frac{\text{Rate}_\text{new}}{\text{Rate}_\text{initial}} = \frac{k \cdot 2^{x - y} [P]^x [Q]^y}{k [P]^x [Q]^y}\)
Cancel out the common terms:
\(\frac{\text{Rate}_\text{new}}{\text{Rate}_\text{initial}} = 2^{x - y}\)
The ratio of the new rate to the initial rate is \(2^{x - y}\).