Question

$A\to 2B$
The initial rates method was used to study the reaction above. Determine the rate expression and calculate the rate constant for the reaction.
$\begin{array}{cc}\left[A\right]\left(mol/L\right)& rate(mol/L.s)\end{array}\begin{array}{c}0.250\\ 0.500\\ 1.00\end{array}\begin{array}{c}3.40\times {10}^2\\ 1.36\times {10}^3\\ 5.44\times {10}^3\end{array}$

• A. Rate = $5.44\times {10}^3{\left[A\right]}^2$
• B. Rate = $1.36\times {10}^3{\left[A\right]}^2$
• C. Rate = $5.44\times {10}^3\left[A\right]$
• D. Rate = $1.36\times {10}^3\left[A\right]$
• E. Rate = $1.84\times {10}^{-4}{\left[A\right]}^2$

Answer 1

The correct option is A Rate = $5.44\times {10}^3{\left[A\right]}^2$

The reaction is given as $A$$\to$$2B$. Let us suppose that the rate constant for this reaction is $k$, and the order of the reaction is $n$. Rate equation is expressed as, $Rate$$=$ $k$$[A{]}^n$, where $k$ is the rate constant and $n$ is the order of the reaction.

From the first set of numbers, we can put them in the equation and say-

$3.40\ast {10}^2$$=$$k$$[0.250{]}^n$.........(1)

From the second set of numbers, we can write-

$1.36\ast {10}^3$$=$$k$$[0.500{]}^n$.........(2)

Dividing, (1) by (2) we get,

$(3.40\ast {10}^2)$$/$$(1.36\ast {10}^3)$$=$$[0.250{]}^n$$/$$[0.500{]}^n$

That is $0.25$$=$$[0.250{]}^n$$/$$2^n$$[0.250{]}^n$, because $(.500{)}^n$ can be written as $2^n$$\ast$$(.25{)}^n$

Or, $0.25$$=$$1$$/$$2^n$

Or, $n$$=$$2$

Thus, the rate will depend on $[A{]}^2$, since $n$ is $2$

Putting the value of $n$ as $2$ in (1) we get,

$3.40\ast {10}^2$$=$$k$$[0.250{]}^2$

From here the value of $k$ is obtained as $5.44\ast {10}^3$.

Thus, the rate expression is -

$Rate$ $=$ $5.44\ast {10}^3$$[A{]}^2$

Hence, option A is the correct answer.