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Question: What is an order with respect to A, B, C respectively? \(\text{ }\left[ \text{A} \right]\text...

What is an order with respect to A, B, C respectively?

 [A] \text{ }\left[ \text{A} \right]\text{ } [C] \text{ }\left[ \text{C} \right]\text{ } [B] \text{ }\left[ \text{B} \right]\text{ }Rate (M /sec)
 0.2 \text{ 0}\text{.2 } 0.1 \text{ 0}\text{.1 } 0.02 \text{ 0}\text{.02 } 0.0 ×10−3 \text{ 0}\text{.0 }\times \text{1}{{\text{0}}^{-3}}\text{ }
 0.1 \text{ 0}\text{.1 } 0.2 \text{ 0}\text{.2 } 0.02 \text{ 0}\text{.02 } 2.01 ×10−3 \text{ 2}\text{.01 }\times \text{1}{{\text{0}}^{-3}}\text{ }
 0.1 \text{ 0}\text{.1 } 1.8 \text{ 1}\text{.8 } 0.18 \text{ 0}\text{.18 } 6.03 ×10−3 \text{ 6}\text{.03 }\times \text{1}{{\text{0}}^{-3}}\text{ }
 0.2 \text{ 0}\text{.2 } 0.1 \text{ 0}\text{.1 } 0.08 \text{ 0}\text{.08 } 6.464 ×10−3 \text{ 6}\text{.464 }\times \text{1}{{\text{0}}^{-3}}\text{ }

A)  −1 , 1 , 32 \text{ }-1\text{ , 1 , }\dfrac{3}{2}\text{ }
B)  −1 , 1 , 12 \text{ }-1\text{ , 1 , }\dfrac{1}{2}\text{ }
C)  1 , 32 ,−1 \text{ }1\text{ , }\dfrac{3}{2}\text{ },-\text{1 }
D)  1 , −1 , 32 \text{ }1\text{ , }-\text{1 , }\dfrac{3}{2}\text{ }

Explanation

Solution

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 R = knCn \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,
 R = k Ax By Cz \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 R = knCn \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,
 R = k Ax By Cz \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,
 0.0×10−3 = k [0.2]x [0.1]y[0.02]z \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,
 2.01×10−3 = k [0.1]x [0.2]y[0.02]z \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,
 6.03×10−3 = k [0.1]x [1.8]y[0.18]z \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,
 6.464×10−3 = k [0.2]x [0.1]y[0.08]z \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,
 2.01×10−3 0.0×10−3=k [0.1]x [0.2]y[0.02]z k [0.2]x [0.1]y[0.02]z  \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,
 22 = [0.1]x[0.2]y  [0.2]x[0.1]y  = (12)x(2)y \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 = x−y 2\text{ = x}-\text{y } (5)
Let’s first divide the equation (1) by the equation (3) we have,
 2.01×10−3 6.03×10−3=k [0.1]x [0.2]y[0.02]z k [0.1]x [1.8]y[0.18]z  \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,
  1 3= [0.2]y[0.02]z  [1.8]y[0.18]z  = (0.21.8)y(0.020.18)z \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,
13=(19)y(19)z \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,

& \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{aligned}$$ Using law of exponents, we have $\text{ }-1\text{ = }-\text{2y }-\text{2z }\Rightarrow \text{ 2y + 2z = 1 }$ (6) Now to determine the values let's divide the equation (1) by the equation (4) .we have, $$\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, $\begin{aligned} & \text{ 2y + 2z = 1 } \\\ & \Rightarrow 2y\text{ + 2 }\left( \dfrac{3}{2} \right)\text{ = 1 } \\\ & \Rightarrow y\text{ = }\dfrac{-2}{2}\text{ = }-1\text{ } \\\ \end{aligned}$ 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, $\begin{aligned} & \text{ }2\text{ = x}-\left( -1 \right) \\\ & \Rightarrow \text{x = 1 } \\\ \end{aligned}$ 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.