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:

\(2^{x - y}\)

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 \(2^{x - y}\).