
What is an order with respect to A, B, C respectively?
$\text{ }\left[ \text{A} \right]\text{ }$ $\text{ }\left[ \text{C} \right]\text{ }$ $\text{ }\left[ \text{B} \right]\text{ }$ Rate (M /sec) $\text{ 0}\text{.2 }$ $\text{ 0}\text{.1 }$ $\text{ 0}\text{.02 }$ $\text{ 0}\text{.0 }\times \text{1}{{\text{0}}^{-3}}\text{ }$ $\text{ 0}\text{.1 }$ $\text{ 0}\text{.2 }$ $\text{ 0}\text{.02 }$ $\text{ 2}\text{.01 }\times \text{1}{{\text{0}}^{-3}}\text{ }$ $\text{ 0}\text{.1 }$ $\text{ 1}\text{.8 }$ $\text{ 0}\text{.18 }$ $\text{ 6}\text{.03 }\times \text{1}{{\text{0}}^{-3}}\text{ }$ $\text{ 0}\text{.2 }$ $\text{ 0}\text{.1 }$ $\text{ 0}\text{.08 }$ $\text{ 6}\text{.464 }\times \text{1}{{\text{0}}^{-3}}\text{ }$
A) $\text{ }-1\text{ , 1 , }\dfrac{3}{2}\text{ }$
B) $\text{ }-1\text{ , 1 , }\dfrac{1}{2}\text{ }$
C) $\text{ }1\text{ , }\dfrac{3}{2}\text{ },-\text{1 }$
D) $\text{ }1\text{ , }-\text{1 , }\dfrac{3}{2}\text{ }$
$\text{ }\left[ \text{A} \right]\text{ }$ | $\text{ }\left[ \text{C} \right]\text{ }$ | $\text{ }\left[ \text{B} \right]\text{ }$ | Rate (M /sec) |
$\text{ 0}\text{.2 }$ | $\text{ 0}\text{.1 }$ | $\text{ 0}\text{.02 }$ | $\text{ 0}\text{.0 }\times \text{1}{{\text{0}}^{-3}}\text{ }$ |
$\text{ 0}\text{.1 }$ | $\text{ 0}\text{.2 }$ | $\text{ 0}\text{.02 }$ | $\text{ 2}\text{.01 }\times \text{1}{{\text{0}}^{-3}}\text{ }$ |
$\text{ 0}\text{.1 }$ | $\text{ 1}\text{.8 }$ | $\text{ 0}\text{.18 }$ | $\text{ 6}\text{.03 }\times \text{1}{{\text{0}}^{-3}}\text{ }$ |
$\text{ 0}\text{.2 }$ | $\text{ 0}\text{.1 }$ | $\text{ 0}\text{.08 }$ | $\text{ 6}\text{.464 }\times \text{1}{{\text{0}}^{-3}}\text{ }$ |
Answer
472.5k+ views
Hint: differential rate expression is a method used to determine the order of reaction with respect to reactant the rate of an nth-order reaction is given by\[\text{ R = }{{\text{k}}_{\text{n}}}{{\text{C}}^{\text{n}}}\text{ }\].where R is the rate of reaction, k is rate constant and C is concentration of reactant. If a reaction involves three reactants let A, B, and C the rate of reaction R is written as,
$\text{ R = k }{{\text{A}}^{\text{x }}}{{\text{B}}^{\text{y}}}\text{ }{{\text{C}}^{\text{z}}}\text{ }$
Where x, y, and z are the order of reaction concerning A, B, and C .here we will apply the differential rate law equation to get the relation between the order. This variable can be obtained by keeping one variable constant.
Complete Solution :
According to the differential rate expression the rate of an nth-order reaction is given by\[\text{ R = }{{\text{k}}_{\text{n}}}{{\text{C}}^{\text{n}}}\text{ }\].where R is the rate of reaction, k is rate constant and C is the concentration of reactant. For a reaction involving three reactants let A, B, and C the rate of reaction R is written as,
$\text{ R = k }{{\text{A}}^{\text{x }}}{{\text{B}}^{\text{y}}}\text{ }{{\text{C}}^{\text{z}}}\text{ }$ (a)
Where x, y, and z are order of reaction with respect to A, B, and C.
We will solve this question using four equations.
Rate reaction for experiment 1 using equation (a) can be written as,
$\text{ 0}\text{.0}\times \text{1}{{\text{0}}^{-3}}\text{ = k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }$ (1)
Rate reaction for experiment 2 using equation (a) can be written as,
$\text{ 2}\text{.01}\times \text{1}{{\text{0}}^{-3}}\text{ = k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 0.2 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }$ (2)
Rate reaction for experiment 3 using equation (a) can be written as,
$\text{ 6}\text{.03}\times \text{1}{{\text{0}}^{-3}}\text{ = k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 1.8 \right]}^{\text{y}}}{{\left[ 0.18 \right]}^{\text{z}}}\text{ }$ (4)
Rate reaction for experiment 4 using equation (a) can be written as,
$\text{ 6}\text{.464}\times \text{1}{{\text{0}}^{-3}}\text{ = k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.08 \right]}^{\text{z}}}\text{ }$ (3)
Part A) Determine relation between variables A, B and C:
We are interested to determine the values of x, y and z .to determine the values lets first divide the equation (1) by the equation (2) .we have,
$\text{ }\dfrac{\text{2}\text{.01}\times \text{1}{{\text{0}}^{-3}}\text{ }}{\text{0}\text{.0}\times \text{1}{{\text{0}}^{-3}}}=\dfrac{\text{k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 0.2 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }}{\text{k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }}\text{ }$
Simply the above equation cancels out common terms from the numerator and denominator. Simplified equation is as shown below,
$\text{ }{{\text{2}}^{\text{2}}}\text{ =}\dfrac{\text{ }{{\left[ \text{0}\text{.1} \right]}^{\text{x}}}{{\left[ \text{0}\text{.2} \right]}^{\text{y}}}\text{ }}{\text{ }{{\left[ \text{0}\text{.2} \right]}^{\text{x}}}{{\left[ \text{0}\text{.1} \right]}^{\text{y}}}\text{ }}\text{ = }{{\left( \dfrac{1}{2} \right)}^{\text{x}}}{{\left( 2 \right)}^{\text{y}}}\text{ }$
Now simply the above relation as follows .we have,
$2\text{ = x}-\text{y }$ (5)
Let’s first divide the equation (1) by the equation (3) we have,
$\text{ }\dfrac{\text{2}\text{.01}\times \text{1}{{\text{0}}^{-3}}\text{ }}{\text{6}\text{.03}\times \text{1}{{\text{0}}^{-3}}}=\dfrac{\text{k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 0.2 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }}{\text{k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 1.8 \right]}^{\text{y}}}{{\left[ 0.18 \right]}^{\text{z}}}\text{ }}\text{ }$
Simply the above equation cancels out common terms from the numerator and denominator. Simplified equation is as shown below,
$\text{ }\dfrac{\text{ 1 }}{\text{3}}\text{=}\dfrac{\text{ }{{\left[ \text{0}\text{.2} \right]}^{\text{y}}}{{\left[ \text{0}\text{.02} \right]}^{\text{z}}}\text{ }}{\text{ }{{\left[ \text{1}\text{.8} \right]}^{\text{y}}}{{\left[ \text{0}\text{.18} \right]}^{\text{z}}}\text{ }}\text{ = }{{\left( \dfrac{\text{0}\text{.2}}{\text{1}\text{.8}} \right)}^{\text{y}}}{{\left( \dfrac{\text{0}\text{.02}}{\text{0}\text{.18}} \right)}^{\text{z}}}\text{ }$
Now simply the above relation as follows .we have,
$\dfrac{1}{3}\text{=}{{\left( \dfrac{1}{9} \right)}^{\text{y}}}{{\left( \dfrac{1}{9} \right)}^{\text{z}}}\text{ }$
On taking reciprocal of the above equation. we have,
\[\begin{align}
& \text{ }{{\text{3}}^{-1}}\text{ = }{{\text{9}}^{-\left( \text{y+z} \right)}}\text{ } \\
& \Rightarrow {{\text{3}}^{-1}}\text{ = }{{\text{3}}^{-2\left( \text{y+z} \right)}}\text{ } \\
\end{align}\]
Using law of exponents, we have
$\text{ }-1\text{ = }-\text{2y }-\text{2z }\Rightarrow \text{ 2y + 2z = 1 }$ (6)
Now to determine the values let's divide the equation (1) by the equation (4) .we have,
\[\text{ }\dfrac{\text{6}\text{.464}\times \text{1}{{\text{0}}^{-3}}\text{ }}{0.0\times \text{1}{{\text{0}}^{-3}}}=\dfrac{\text{k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.08 \right]}^{\text{z}}}\text{ }}{\text{k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }}\text{ }\]
Simply the above equation cancels out common terms from the numerator and denominator. Simplified equation is as shown below,
${{\text{2}}^{\text{3}}}=\dfrac{\text{ }{{\left[ \text{0}\text{.2} \right]}^{\text{x}}}{{\left[ \text{0}\text{.1} \right]}^{\text{y}}}{{\left[ \text{0}\text{.02} \right]}^{\text{z}}}\text{ }}{\text{ }{{\left[ \text{0}\text{.2} \right]}^{\text{x}}}{{\left[ \text{0}\text{.1} \right]}^{\text{y}}}{{\left[ \text{0}\text{.08} \right]}^{\text{z}}}\text{ }}\text{ = }{{\left( \dfrac{\text{0}\text{.08}}{\text{0}\text{.02}} \right)}^{\text{z}}}\text{ }$
Now simply the above relation as follows .we have,
\[{{\text{2}}^{\text{3}}}\text{=}{{\left( {{2}^{2}} \right)}^{\text{z}}}\text{ }\]
$\text{ 3 = 2z }\Rightarrow \text{ z = }\dfrac{3}{2}\text{ }$
Let's substitute the value of z in equation (6) we have,
$\begin{align}
& \text{ 2y + 2z = 1 } \\
& \Rightarrow 2y\text{ + 2 }\left( \dfrac{3}{2} \right)\text{ = 1 } \\
& \Rightarrow y\text{ = }\dfrac{-2}{2}\text{ = }-1\text{ } \\
\end{align}$
Part B) Determine order of reaction with respect to reactant:
Let's substitute the value of y in equation (5) we get the value of x as,
$\begin{align}
& \text{ }2\text{ = x}-\left( -1 \right) \\
& \Rightarrow \text{x = 1 } \\
\end{align}$
Thus the order of reactant with respect to A, B, and C is 1, $-1$ and $\left( \dfrac{3}{2} \right)$ respectively.
So, the correct answer is “Option D”.
Note: Remember that to simply the above relation we are using the law of exponents. According to which if bases are equal then exponents are also equal i.e. $\text{ }{{\text{A}}^{\text{x}}}\text{ = }{{\text{A}}^{\text{y}}}\text{ }\Rightarrow \text{ x = y }$ .Figure out a base then equate the exponents to get the order or reaction with respect to reactant.
$\text{ R = k }{{\text{A}}^{\text{x }}}{{\text{B}}^{\text{y}}}\text{ }{{\text{C}}^{\text{z}}}\text{ }$
Where x, y, and z are the order of reaction concerning A, B, and C .here we will apply the differential rate law equation to get the relation between the order. This variable can be obtained by keeping one variable constant.
Complete Solution :
According to the differential rate expression the rate of an nth-order reaction is given by\[\text{ R = }{{\text{k}}_{\text{n}}}{{\text{C}}^{\text{n}}}\text{ }\].where R is the rate of reaction, k is rate constant and C is the concentration of reactant. For a reaction involving three reactants let A, B, and C the rate of reaction R is written as,
$\text{ R = k }{{\text{A}}^{\text{x }}}{{\text{B}}^{\text{y}}}\text{ }{{\text{C}}^{\text{z}}}\text{ }$ (a)
Where x, y, and z are order of reaction with respect to A, B, and C.
We will solve this question using four equations.
Rate reaction for experiment 1 using equation (a) can be written as,
$\text{ 0}\text{.0}\times \text{1}{{\text{0}}^{-3}}\text{ = k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }$ (1)
Rate reaction for experiment 2 using equation (a) can be written as,
$\text{ 2}\text{.01}\times \text{1}{{\text{0}}^{-3}}\text{ = k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 0.2 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }$ (2)
Rate reaction for experiment 3 using equation (a) can be written as,
$\text{ 6}\text{.03}\times \text{1}{{\text{0}}^{-3}}\text{ = k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 1.8 \right]}^{\text{y}}}{{\left[ 0.18 \right]}^{\text{z}}}\text{ }$ (4)
Rate reaction for experiment 4 using equation (a) can be written as,
$\text{ 6}\text{.464}\times \text{1}{{\text{0}}^{-3}}\text{ = k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.08 \right]}^{\text{z}}}\text{ }$ (3)
Part A) Determine relation between variables A, B and C:
We are interested to determine the values of x, y and z .to determine the values lets first divide the equation (1) by the equation (2) .we have,
$\text{ }\dfrac{\text{2}\text{.01}\times \text{1}{{\text{0}}^{-3}}\text{ }}{\text{0}\text{.0}\times \text{1}{{\text{0}}^{-3}}}=\dfrac{\text{k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 0.2 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }}{\text{k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }}\text{ }$
Simply the above equation cancels out common terms from the numerator and denominator. Simplified equation is as shown below,
$\text{ }{{\text{2}}^{\text{2}}}\text{ =}\dfrac{\text{ }{{\left[ \text{0}\text{.1} \right]}^{\text{x}}}{{\left[ \text{0}\text{.2} \right]}^{\text{y}}}\text{ }}{\text{ }{{\left[ \text{0}\text{.2} \right]}^{\text{x}}}{{\left[ \text{0}\text{.1} \right]}^{\text{y}}}\text{ }}\text{ = }{{\left( \dfrac{1}{2} \right)}^{\text{x}}}{{\left( 2 \right)}^{\text{y}}}\text{ }$
Now simply the above relation as follows .we have,
$2\text{ = x}-\text{y }$ (5)
Let’s first divide the equation (1) by the equation (3) we have,
$\text{ }\dfrac{\text{2}\text{.01}\times \text{1}{{\text{0}}^{-3}}\text{ }}{\text{6}\text{.03}\times \text{1}{{\text{0}}^{-3}}}=\dfrac{\text{k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 0.2 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }}{\text{k }{{\left[ 0.1 \right]}^{\text{x }}}{{\left[ 1.8 \right]}^{\text{y}}}{{\left[ 0.18 \right]}^{\text{z}}}\text{ }}\text{ }$
Simply the above equation cancels out common terms from the numerator and denominator. Simplified equation is as shown below,
$\text{ }\dfrac{\text{ 1 }}{\text{3}}\text{=}\dfrac{\text{ }{{\left[ \text{0}\text{.2} \right]}^{\text{y}}}{{\left[ \text{0}\text{.02} \right]}^{\text{z}}}\text{ }}{\text{ }{{\left[ \text{1}\text{.8} \right]}^{\text{y}}}{{\left[ \text{0}\text{.18} \right]}^{\text{z}}}\text{ }}\text{ = }{{\left( \dfrac{\text{0}\text{.2}}{\text{1}\text{.8}} \right)}^{\text{y}}}{{\left( \dfrac{\text{0}\text{.02}}{\text{0}\text{.18}} \right)}^{\text{z}}}\text{ }$
Now simply the above relation as follows .we have,
$\dfrac{1}{3}\text{=}{{\left( \dfrac{1}{9} \right)}^{\text{y}}}{{\left( \dfrac{1}{9} \right)}^{\text{z}}}\text{ }$
On taking reciprocal of the above equation. we have,
\[\begin{align}
& \text{ }{{\text{3}}^{-1}}\text{ = }{{\text{9}}^{-\left( \text{y+z} \right)}}\text{ } \\
& \Rightarrow {{\text{3}}^{-1}}\text{ = }{{\text{3}}^{-2\left( \text{y+z} \right)}}\text{ } \\
\end{align}\]
Using law of exponents, we have
$\text{ }-1\text{ = }-\text{2y }-\text{2z }\Rightarrow \text{ 2y + 2z = 1 }$ (6)
Now to determine the values let's divide the equation (1) by the equation (4) .we have,
\[\text{ }\dfrac{\text{6}\text{.464}\times \text{1}{{\text{0}}^{-3}}\text{ }}{0.0\times \text{1}{{\text{0}}^{-3}}}=\dfrac{\text{k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.08 \right]}^{\text{z}}}\text{ }}{\text{k }{{\left[ 0.2 \right]}^{\text{x }}}{{\left[ 0.1 \right]}^{\text{y}}}{{\left[ 0.02 \right]}^{\text{z}}}\text{ }}\text{ }\]
Simply the above equation cancels out common terms from the numerator and denominator. Simplified equation is as shown below,
${{\text{2}}^{\text{3}}}=\dfrac{\text{ }{{\left[ \text{0}\text{.2} \right]}^{\text{x}}}{{\left[ \text{0}\text{.1} \right]}^{\text{y}}}{{\left[ \text{0}\text{.02} \right]}^{\text{z}}}\text{ }}{\text{ }{{\left[ \text{0}\text{.2} \right]}^{\text{x}}}{{\left[ \text{0}\text{.1} \right]}^{\text{y}}}{{\left[ \text{0}\text{.08} \right]}^{\text{z}}}\text{ }}\text{ = }{{\left( \dfrac{\text{0}\text{.08}}{\text{0}\text{.02}} \right)}^{\text{z}}}\text{ }$
Now simply the above relation as follows .we have,
\[{{\text{2}}^{\text{3}}}\text{=}{{\left( {{2}^{2}} \right)}^{\text{z}}}\text{ }\]
$\text{ 3 = 2z }\Rightarrow \text{ z = }\dfrac{3}{2}\text{ }$
Let's substitute the value of z in equation (6) we have,
$\begin{align}
& \text{ 2y + 2z = 1 } \\
& \Rightarrow 2y\text{ + 2 }\left( \dfrac{3}{2} \right)\text{ = 1 } \\
& \Rightarrow y\text{ = }\dfrac{-2}{2}\text{ = }-1\text{ } \\
\end{align}$
Part B) Determine order of reaction with respect to reactant:
Let's substitute the value of y in equation (5) we get the value of x as,
$\begin{align}
& \text{ }2\text{ = x}-\left( -1 \right) \\
& \Rightarrow \text{x = 1 } \\
\end{align}$
Thus the order of reactant with respect to A, B, and C is 1, $-1$ and $\left( \dfrac{3}{2} \right)$ respectively.
So, the correct answer is “Option D”.
Note: Remember that to simply the above relation we are using the law of exponents. According to which if bases are equal then exponents are also equal i.e. $\text{ }{{\text{A}}^{\text{x}}}\text{ = }{{\text{A}}^{\text{y}}}\text{ }\Rightarrow \text{ x = y }$ .Figure out a base then equate the exponents to get the order or reaction with respect to reactant.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

Arrange Water ethanol and phenol in increasing order class 11 chemistry CBSE

Name the nuclear plant located in Uttar Pradesh class 11 social science CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

A mixture of o nitrophenol and p nitrophenol can be class 11 chemistry CBSE
