Introduction to rate of reaction - class-X

The term $-\dfrac{dc}{dt}$ in a rate expression indicates the decrease in concentration of the reactant with time. It is the minus sign which is used to show the decrease in concentration of the reactant.

K$= \frac { 2.303 }{ t } \log { \left[ \dfrac { { A } _{ 0 } }{ A }  \right]  } $

A] True. for first order reactions, products are formed at the fastest rate during the  beginning of the reaction.

Given  rate is $k[A]{ [B] }^{ 2 }$

• Temperature is a measure of the kinetic energy of a system, so higher temperature implies higher average kinetic energy of molecules and more collisions per unit time. Hence an increase in temperature is accompanied by an increase in the reaction rate. 
  
• A higher concentration of reactants leads to more effective collisions per unit time, which leads to an increased reaction rate (except for zero order reactions). Similarly, a higher concentration of products tends to be associated with a lower reaction rate.
• Catalysts (e.g., enzymes) lower the activation energy of a chemical reaction and increase the rate of a chemical reaction without being consumed in the process. 
  
• Hence, using a positive catalyst and increasing the temperature, both can enhance the rate of reaction.

Enzymes are the very efficient catalyst for the enzymatic biochemical reaction and Enzymes speed up the reaction by providing lower activation energy.

According to the balanced chemical equation 2SO2 + O2 -> 2SO3, the stoichiometric coefficients for SO2 and SO3 are both 2, meaning they react and form in a 1:1 molar ratio. Since their molar masses are different, we account for them: rate of SO2 consumption = rate of SO3 formation * (Molar Mass of SO2 / Molar Mass of SO3). Molar mass of SO2 is 64 g/mol and SO3 is 80 g/mol. Thus, rate = 1.6 * 10^-3 * (64 / 80) = 1.28 * 10^-3 kg min^-1.

For the reaction C4H8 -> 2C2H4, if the ratio [C2H4]/[C4H8] = 1, then [C2H4] = [C4H8]. Using the integrated rate law for a first-order reaction, we solve for t.

Rate$=$ ${K _{{p^2}}}{H _2}S \times P{O _2} = x$ $($rate$)$

$A+B+C\longrightarrow D+E$

The slowest step in the mechanism of any reaction is the rate-determining step of the reaction.

The rate of reaction depends upon the nature of reactants, the concentration of the reactants, temperature, and catalyst.

So, the correct option is $D$

(i)   Addition of Helium gas will not increase the rate of reaction. It will either remain same or decrease depending upon 

A change in pressure effects only if there is a difference between the number of moles of gaseous products & the number of moles of gaseous reactant. In short, a change in pressure affects only if there is at least one gas either in products or reactants. 

Reaction could be such as:

Molecularity is a theoretical concept describing the mechanism of an elementary step, whereas the rate of a reaction is an experimental observation influenced by external factors like temperature, concentration, and catalysts.

By unit of rate constant it is clear,that the $reaction$ is zero order. 

Given that

The given reaction is $2A+B\longrightarrow C$

The given reaction is $A+2B\longrightarrow2C+O$

Decomposition of N2O5 is a known first-order reaction. The rate constant can be determined from the time-volume data using the first-order integrated rate equation.

The given reaction is:-

Its a 1st order reaction,

The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed. Definition of a rate of reaction. We, in general, use a negative sign for reactants because its concentration is decreasing as the chemical reaction is taking place.

$Rate = -\dfrac {1}{2} \dfrac {\triangle [X]}{\triangle t}$
$= -\dfrac {1}{2} \dfrac {(3 - 2)}{5} = -0.1\ mol\ L^{-1} min^{-1}$
Negative sign signifies the decrease in concentration.

Rate of reaction- The speed at which a chemical reaction proceeds, It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or the concentration of a reactant that is consumed in a unit of time.

For the reaction,

The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed. Hence rate of reaction is inversely related as the time increases rate of reaction decreases. Hence given statement is true.

$Rate = k[A]^{2} [B]^{3} = a$

Rate of reaction $= \dfrac {\triangle [X]}{\triangle t}$

$r= K[A]^{o} $ $[B]^3$

$ \implies \space \space \space A(g) \space \space \space \space \space \space+ \space \space \space \space \space \space 2B(g) \rightarrow \space \space \space \space \space \space \space \space C(g) + D(g) $
$ At \space t=0 \space0.6 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 0.8 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 0 \space \space \space \space \space \space \space \space \space \space \space \space \space 0 $
$ At \space t=t \space (0.6-0.2) \space \space \space \space (0.8 - 2 \times 0.2) \space \space 0.2 \space \space \space \space \space \space \space \space \space \space0.2$

$ Rate _i = [0.6][0.8]^2$
$ Rate _t = [0.4][0.4]^2$
$ \dfrac{Rate _t}{Rate _i} = \dfrac{1}{6}$

Time does not affect the rate of a reaction. Hence, in the rate law expression $ \displaystyle rate = k[A]^n$, there is no term for time.
Nature of the reactants, Concentrations, surface area exposed and temperature affects  the rate of a reaction.

Change in pressure $=120-100=20 mmHg$

If, $Rate=kx[A]^2[B]^1$

$order=3$.

If pressure is increased by factor of $2$, then rate will be increased by factor of $2^3=8$.

$\therefore $ Rate becomes eight times.

The given reaction is $A+B\longrightarrow$ Product