Question

Consider the kinetic data given in the following table for the reaction:
\[A+B+C\to product\]

Experiment No.	[A]$mol\text{ }d{{m}^{-3}}$	[B]$mol\text{ }d{{m}^{-3}}$	[C]$mol\text{ }d{{m}^{-3}}$	Rate of reaction$mol\text{ }d{{m}^{-3}}{{s}^{-1}}$
1	0.2	0.1	0.1	$6.0\times {{10}^{-5}}$
2	0.2	0.2	0.1	$6.0\times {{10}^{-5}}$
3	0.2	0.1	0.2	$1.2\times {{10}^{-4}}$
4	0.3	0.1	0.1	$9.0\times {{10}^{-5}}$

The rate of reaction for the concentrations [A] = $0.15\text{ }mol\text{ }d{{m}^{-3}}$, [B] = $0.25\text{ }mol\text{ }d{{m}^{-3}}$, and [C] = $0.15\text{ }mol\text{ }d{{m}^{-3}}$ is found to be $Y\times {{10}^{-5}}\text{ }mol\text{ }d{{m}^{-3}}{{s}^{-1}}$. What is the value of $Y$? Give your answer in the nearest integer value.

Answer 1

Hint: Recall the basic formula that is given for the rate of the reaction which includes the concentration of the reactants and their stoichiometric coefficients in the reaction. It also involves the rate constant of the reaction.

Complete step by step solution:
We know that the most general formula that we use to determine the rate of a reaction includes the product of the concentrations of all the reactants present in the reaction. Each concentration is raised to a power that is equal to the stoichiometric coefficient. Consider the reaction:
\[aA+bB\to cC\]
Where, A and B are the reactants and a and b are their stoichiometric coefficients. The corresponding expression for the calculation of the rate for this reaction will be:
\[r=k{{[A]}^{a}}{{[B]}^{b}}\]
According to this rule, while assuming that the stoichiometric coefficients for A, B, and C in the reaction given in the questions as a, b, and c, the expression for the rate will be:
\[r=k{{[A]}^{a}}{{[B]}^{b}}{{[C]}^{c}}\]
We will determine the values of a, b, and c based on the information of the 4 reactions that is given to us. First, we will prepare the rate equations for each of the given reactions and label them as i), ii), iii), and iv).
\[\begin{align}
& i)6.0\times {{10}^{-5}}=k{{(0.2)}^{a}}{{(0.1)}^{b}}{{(0.1)}^{c}} \\
& ii)6.0\times {{10}^{-5}}=k{{(0.2)}^{a}}{{(0.2)}^{b}}{{(0.1)}^{c}} \\
& iii)1.2\times {{10}^{-4}}=k{{(0.2)}^{a}}{{(0.1)}^{b}}{{(0.2)}^{c}} \\
& iv)9.0\times {{10}^{-5}}=k{{(0.3)}^{a}}{{(0.1)}^{b}}{{(0.1)}^{c}} \\
\end{align}\]
We will now divide equation i) by ii), iii), and iv) to get the values of b, c, and a respectively.
- For b, by the laws of indices
\[\begin{align}
& \dfrac{6.0\times {{10}^{-5}}}{6.0\times {{10}^{-5}}}=\dfrac{k{{(0.2)}^{a}}{{(0.1)}^{b}}{{(0.1)}^{c}}}{k{{(0.2)}^{a}}{{(0.2)}^{b}}{{(0.1)}^{c}}} \\
& 1={{\left( \dfrac{0.1}{0.2} \right)}^{b}} \\
& b=0 \\
\end{align}\]
- For c, by the laws of indices
\[\begin{align}
& \dfrac{6.0\times {{10}^{-5}}}{1.2\times {{10}^{-4}}}=\dfrac{k{{(0.2)}^{a}}{{(0.1)}^{b}}{{(0.1)}^{c}}}{k{{(0.2)}^{a}}{{(0.1)}^{b}}{{(0.2)}^{c}}} \\
& \dfrac{1}{2}={{\left( \dfrac{0.1}{0.2} \right)}^{c}} \\
& c=1 \\
\end{align}\]
- For a, by the laws of indices
\[\begin{align}
& \dfrac{6.0\times {{10}^{-5}}}{9.0\times {{10}^{-5}}}=\dfrac{k{{(0.2)}^{a}}{{(0.1)}^{b}}{{(0.1)}^{c}}}{k{{(0.3)}^{a}}{{(0.1)}^{b}}{{(0.1)}^{c}}} \\
& \dfrac{2}{3}={{\left( \dfrac{0.2}{0.3} \right)}^{a}} \\
& a=1 \\
\end{align}\]
Now that we have the stoichiometric coefficients we can substitute them along with the given concentrations of the reactants and find the rate of the reaction in terms of $Y$. But we first have to find the value of $k$, which is the rate constant of the reaction, it will remain the same no matter what the concentration of the reactants is. To find that, we will put the values of a, b, and c in equation i) and we will get:
\[\begin{align}
& 6.0\times {{10}^{-5}}=k{{(0.2)}^{1}}{{(0.1)}^{0}}{{(0.1)}^{1}} \\
& k=\dfrac{6.0\times {{10}^{-5}}}{0.2\times 0.1} \\
& k=3.0\times {{10}^{-3}} \\
\end{align}\]
Now, we will put all these values in the expression for the rate of a reaction and find the rate under the given concentrations of the reactants. The equation will be as follows:
\[\begin{align}
& r=3.0\times {{10}^{-3}}{{(0.15)}^{1}}{{(0.25)}^{0}}{{(0.15)}^{1}} \\
& r=3.0\times {{10}^{-3}}\times 2.25\times {{10}^{-2}} \\
& r=6.75\times {{10}^{-5}} \\
& r\cong 7\times {{10}^{-5}} \\
\end{align}\]

So, the value of $Y$ will be 7 for this reaction under the given concentrations of the reactants.

Note: It sometimes happens that the power to which we raise the concentrations of the reactants is not even remotely related to the stoichiometric coefficients. For example, here, the coefficient of B reactant is 0, this implies that this reactant is not essential to the reaction which may or may not be true. But we need not take this fact into consideration in this problem since we are calculating the power from the rats and not the reaction.