Tag: dependence of reaction rate on concentration of reactants

Quizzes (0) Questions (85)

Questions Related to dependence of reaction rate on concentration of reactants

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

The rate constant (K) for the reaction, $2A+B\rightarrow$ Product was found to be $2.5\times 10^{-5}$ litre $mol^{-1} sec^{-1}$ after 15 sec, $2.60\times 10^{-5} litre\  mol^{-1} sec^{-1}$ after 30 sec and $2.55\times 10^{-5} litre \ mol^{-1}sec^{-1}$ after 50 sec. The order of reaction is:

  1. $2$
  2. $3$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 The order of reaction is 2.


K does not change with time; thus its value remains unchanged during the reaction.

By seeing the equation, it can be said that reaction is a third order but the unit of K suggest it to be 2nd order.

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

Assertion : In a second-order reaction with respect to A, when you double [A], the rate is quadrupled.
Reason : The rate equation is $\displaystyle r={ k\left[ A \right]  }^{ 2 }$ for such a reaction.

  1. <p>Both Assertion and Reason are true and Reason is the correct explanation of Assertion</p>

  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

  3. Assertion is true but Reason is false

  4. Assertion is false but Reason is true

  5. Both Assertion and Reason are false
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rate equation is as follows:
$Older Rate = k [A]^{2}$                       equation -1
This shows that reaction is $2nd $ order.
So if we double the concentration of the A , then new reaction rate can be written as follows:
$New Rate = k [2A]^{2}$
$New Rate = k\times 4[A]^{2}$
$New Rate = 4 k [A]^{2}$                     equation -2
Now, comparing equation -1 and 2 we get :
New Rate = $4\times Older Rate$


Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

For a gaseous reaction, $A\left( g \right) \longrightarrow $ Product, which one of the following is correct relation among $\dfrac { dP }{ dt } ,\dfrac { dn }{ dt }$ and $\dfrac { dc }{ dt } $?
($\dfrac { dP }{ dt } =$ Rate of reaction in $atm$ ${ sec }^{ -1 }$; $\dfrac { dc }{ dt } =$ Rate of reaction in molarity ${ sec }^{ -1 }$; $\dfrac { dn }{ dt } =$ Rate of reaction in $mol$ ${ sec }^{ -1 }$)

  1. $\dfrac { dc }{ dt } =\dfrac { dn }{ dt } =-\dfrac { dP }{ dt } $
  2. $-\dfrac { dc }{ dt } =-\dfrac { 1 }{ V } \dfrac { dn }{ dt } =-\dfrac {1}{RT}\dfrac{ dP }{ dt } $
  3. $\dfrac { dc }{ dt } =\dfrac { V }{ RT } \dfrac { dn }{ dt } =\dfrac { dP }{ dt } $

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$PV=nRT$
$\frac{dP}{dt}.V=RT.\frac{dn}{dt}$=> $\frac{1}{V}\frac{dn}{dt}=\frac{1}{RT}\frac{dP}{dt}$-(1)
Now, PV=nRT or $P=\frac{n}{V}RT$
Concentration $c=\frac{n}{V}$
P=cRT
$\frac{dP}{dt}=\frac{dc}{dt}.RT$=> \frac{dc}{dt}=\frac{1}{RT}\frac{dP}{dt}$
$-\frac{dc}{dt}=-\frac{1}{V}\frac{dn}{dt}=-\frac{1}{RT}\frac{dP}{dt}$
Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

The rate constant for the reaction is $2 10^{-4} s^{-1}.$ The reaction is :

  1. First order

  2. Second order

  3. Third order

  4. Zero order
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The unit of the rate constant is s^-1. This unit is characteristic of a first-order reaction, where the rate constant k has dimensions of time^-1.

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

For a second-order reaction of the type rate=$k[A]^2$ the plot of $\dfrac{1}{[A] _1}$ versus t is linear with a: 

  1. positive slope and zero intercept

  2. positive slope and non-zero intercept

  3. negative slope and zero intercept

  4. negative slope and non-zero intercept
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that for a second order reaction

$Rate= K[A]^2$

We know that rate of the reaction is decrease in the concentration of the reactant with respect to time.

$Rate= \cfrac {-d[A]}{dt}$
$\Rightarrow$ $\cfrac {-d[A]}{dt}=K[A]^2$
$\Rightarrow \cfrac {-d[A]}{[A]^2}=+Kdt$

Integrating both sides we have:-
$\Rightarrow -\int \cfrac {d[A]}{[A]^2}=+\int Kdt$
$\Rightarrow - \left[\cfrac {-1}{[A]}\right]=+Kt+C$
$\Rightarrow Kt+C= \cfrac {1}{[A]}$      $- (i)$

Now, from the equation $(i)$ we can see that graph of $\cfrac {1}{[A]}$ $V/s$ $t$ has a positive slope $K$ and non-zero intercept $C$.