
For the chemical reaction \[x\to y\], it is found that the rate of reaction increases 2.25 times when the concentration of ‘x’ is increased 1.5 times. What is the order of reaction?
a.) One
b.) Two
c.) Three
d.) Zero
Answer
595.2k+ views
Hint: For finding the order of reaction we have to write the rate of reaction for the given reaction by assuming ‘n’ as the order of the reaction, because the change in rate of reaction and concentration are given.
Complete step by step solution:
We know that the order of reaction refers to the power dependence of the rate on the concentration of each reactant. For example a first-order reaction, the rate is dependent on the concentration of a single species.
In the question given that the reaction is:
\[x\to y\]
And also given that, if the rate of reaction increases 2.25 times then the concentration of ‘x’ is increased 1.5 times.
For the given equation let the rate of reaction is:
\[r=k{{[c]}^{n}}\]
Where, r is the rate of reaction,
‘k’ is the rate constant,
‘c’ is the concentration of ‘x’,
‘n’ is the order of the reaction.
\[{{r}_{1}}=k{{[{{c}_{1}}]}^{n}}\]
And,
\[{{r}_{2}}=k{{[{{c}_{2}}]}^{n}}\]
Now by dividing:
\[\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{{{[{{c}_{2}}]}^{n}}}{{{[{{c}_{2}}]}^{n}}}\]
And given that
\[\dfrac{{{r}_{1}}}{{{r}_{2}}}\]= (1/2.25)
And
\[\dfrac{[{{c}_{2}}]}{[{{c}_{2}}]}\]= (1/1.5)
now,
1/(2.25) = \[{{(1/1.5)}^{n}}\]
So, n = 2
We can say that the order of reaction is ‘2’ and the correct answer is “B”.
Note: Here molecularity is not equal to the order of the reaction. And note that here in this question the change in concentration and change in reaction rate are given, not the direct rate of reaction and concentration.
Complete step by step solution:
We know that the order of reaction refers to the power dependence of the rate on the concentration of each reactant. For example a first-order reaction, the rate is dependent on the concentration of a single species.
In the question given that the reaction is:
\[x\to y\]
And also given that, if the rate of reaction increases 2.25 times then the concentration of ‘x’ is increased 1.5 times.
For the given equation let the rate of reaction is:
\[r=k{{[c]}^{n}}\]
Where, r is the rate of reaction,
‘k’ is the rate constant,
‘c’ is the concentration of ‘x’,
‘n’ is the order of the reaction.
\[{{r}_{1}}=k{{[{{c}_{1}}]}^{n}}\]
And,
\[{{r}_{2}}=k{{[{{c}_{2}}]}^{n}}\]
Now by dividing:
\[\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{{{[{{c}_{2}}]}^{n}}}{{{[{{c}_{2}}]}^{n}}}\]
And given that
\[\dfrac{{{r}_{1}}}{{{r}_{2}}}\]= (1/2.25)
And
\[\dfrac{[{{c}_{2}}]}{[{{c}_{2}}]}\]= (1/1.5)
now,
1/(2.25) = \[{{(1/1.5)}^{n}}\]
So, n = 2
We can say that the order of reaction is ‘2’ and the correct answer is “B”.
Note: Here molecularity is not equal to the order of the reaction. And note that here in this question the change in concentration and change in reaction rate are given, not the direct rate of reaction and concentration.
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