${{[{{A}_{2}}]}_{0}}$ ${{[{{B}_{2}}]}_{0}}$ Initial rate of appearance of AB(g) ( in $M{{s}^{-1}}$) 0.10 0.10 $2.5\times {{10}^{-4}}$ 0.20 0.10 $5\times {{10}^{-4}}$ 0.20 0.20 $10\times {{10}^{-4}}$

1)The value of rate constant for the written reaction is:

A. $2.5\times {{10}^{-4}}$

B. $2.5\times {{10}^{-2}}$

C. $1.25\times {{10}^{-2}}$

D. none of these

2)Above given question, the value of rate constant for appearance of AB(g) is:

A. $2.5\times {{10}^{-4}}$

B. $2.5\times {{10}^{-2}}$

C. $1.25\times {{10}^{-2}}$

D. none of these

${{[{{A}_{2}}]}_{0}}$ | ${{[{{B}_{2}}]}_{0}}$ | Initial rate of appearance of AB(g) ( in $M{{s}^{-1}}$) |

0.10 | 0.10 | $2.5\times {{10}^{-4}}$ |

0.20 | 0.10 | $5\times {{10}^{-4}}$ |

0.20 | 0.20 | $10\times {{10}^{-4}}$ |

Answer

Verified

197.1k+ views

**Hint:**These types of questions can be solved by writing the rate equation for the reaction. Then put the values of the concentrations and divide the equations to get the value of the stoichiometric power. Substitute these powers back in the equation again to obtain the rate constant.

**Complete step by step solution:**

1) Reaction rate, in chemistry,is referred to as the speed at which a chemical action proceeds. It is often expressed in terms of either the concentration i.e amount per unit volume of a product that's formed during a unit of your time or the concentration of reactant that's consumed in an infinitesimally small unit of time. Alternatively, it's going to be defined in terms of the amounts of the reactants consumed or products formed in an exceedingly unit of time.

In order to proceed with our question, let us write the rate law equation. We have

$Rate=k{{[{{A}_{0}}]}^{x}}{{[{{B}_{0}}]}^{y}}$, where k is rate constant and x and y are the orders of the reactants

Now from the table, we can write the necessary equations:

$\begin{align}

& k{{[0.1]}^{x}}{{[0.1]}^{y}}=2.5\times {{10}^{-4}}..........(1) \\

& k{{[0.2]}^{x}}{{[0.1]}^{y}}=5\times {{10}^{-4}}.............(2) \\

& k{{[0.2]}^{x}}{{[0.2]}^{y}}=10\times {{10}^{-4}}...........(3) \\

\end{align}$

Now, in order to find the value of the orders x and y, we divide equation (1) by (2) once and (2) by (3). So,

On dividing (1) by (2), we get the value of x=1

On dividing (2) by (3), we get the value of y=1

So, we obtain rate equation as:

$k[0.2][0.1]=5\times {{10}^{-4}}$, by putting the values in equation (2)

On solving, we have

$\begin{align}

& k=\dfrac{5}{2}\times {{10}^{-2}} \\

& k=2.5\times {{10}^{-2}} \\

\end{align}$

So, our answer comes out to be option B.

2) In order to find out the rate constant for appearance for AB(g), we will use the same method as done in the previous question, the answer comes out to be $k=2.5\times {{10}^{-2}}$. The rate constant of the equation is equal to the rate constant of appearance of AB.

So, we get option B as the answer again like in the previous question.

**NOTE:**There is an alternative solution to this question. We can see that when the value of $[{{A}_{2}}]$ is doubled, the rate of appearance is also doubled, and it is the same for $[{{B}_{2}}]$.So even without solving and dividing the equations, we can predict that each of the reactants have an order of 1.

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