To determine how the rate of reaction changes when the concentrations of both reactants are doubled, let's analyze the given rate equation step-by-step:
The rate equation is given by:
[tex]\[ \text{Rate} = K [X]^{5/3} [Y]^{4/3} \][/tex]
Here, [tex]\( K \)[/tex] is the rate constant, [tex]\([X]\)[/tex] is the concentration of reactant X, and [tex]\([Y]\)[/tex] is the concentration of reactant Y.
Now, let's consider what happens when the concentrations of both reactants are doubled. If the original concentrations are [tex]\([X]\)[/tex] and [tex]\([Y]\)[/tex], doubling these concentrations means they become [tex]\([2X]\)[/tex] and [tex]\([2Y]\)[/tex] respectively.
We substitute these new concentrations into the rate equation to find the new rate:
[tex]\[ \text{New Rate} = K [2X]^{5/3} [2Y]^{4/3} \][/tex]
Next, we need to simplify the expression. Applying the exponents individually to the concentration terms:
[tex]\[ [2X]^{5/3} = 2^{5/3} [X]^{5/3} \][/tex]
[tex]\[ [2Y]^{4/3} = 2^{4/3} [Y]^{4/3} \][/tex]
Therefore, the new rate can be written as:
[tex]\[ \text{New Rate} = K (2^{5/3} [X]^{5/3}) (2^{4/3} [Y]^{4/3}) \][/tex]
Combining the terms involving powers of 2:
[tex]\[ \text{New Rate} = K \cdot 2^{5/3} \cdot 2^{4/3} \cdot [X]^{5/3} \cdot [Y]^{4/3} \][/tex]
Since [tex]\( 2^{5/3} \cdot 2^{4/3} = 2^{(5/3 + 4/3)} = 2^3 \)[/tex]:
[tex]\[ \text{New Rate} = K \cdot 2^3 \cdot [X]^{5/3} \cdot [Y]^{4/3} \][/tex]
We know that:
[tex]\[ 2^3 = 8 \][/tex]
Thus, the equation simplifies to:
[tex]\[ \text{New Rate} = 8 \cdot K [X]^{5/3} [Y]^{4/3} \][/tex]
The original rate is:
[tex]\[ \text{Rate} = K [X]^{5/3} [Y]^{4/3} \][/tex]
Comparing the new rate with the original rate, we see that the new rate is 8 times greater than the original rate.
Therefore, if the concentrations of both reactants are doubled, the rate of reaction increases eight times.
The correct answer is:
(c) eight times
