Courses
Courses for Kids
Free study material
Offline Centres
More
Store

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) $\text{ }1\text{ , }-\text{1 , }\dfrac{3}{2}\text{ }$

Last updated date: 21st Jun 2024
Total views: 394.8k
Views today: 8.94k
Verified
394.8k+ 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.