Chemistry Chemical Kinetics

Endothermic reaction is a reaction which absorbs heat energy from the surrounding, making the environment cooler. The products have higher enthalpy than the reactants in this reaction.

On the other hand, Exothermic reaction is a reaction which releases heat energy into the surrounding, making the environment warmer. The products will have lower enthalpy than the reactants in this reaction.

This happens because although they have the same temeprature constant, some of their other properties can also vary such as activation energies, surface area of the reaction, concentration of the reactant, etc. As a result, the rate of a reaction comes out to be different for the reactions because of their dependence on these factors.

This is the type of reaction which bheavies like a first order reaction inspite of being a higher order reaction (second or third). This happens due to a particular reactant being present in an excessive quantity thorughout the chemical reaction.

These are the important highlights of integrated rate equations:

• predict the rate of a reaction
• calculate order of reaction (zero, first, second)
• understand mechanism of a reaction
• compute half life
• calculate the value of k

Yes, the phase of the reactant (solid, liquid or gas) actually affects the rate of a reaction. Gases and Liquids tend to move more freely and effectively as compared to solids, which increaes the frequency of particles colliding with each other.

This is possible because these catalysts have a lower activation energy, which allows a comparatively higher number of particles to react at a faster rate. More number of particles will mean more frequent collisions which speeds up the process.

lnK = -E_a/RT + I
-E_a/R = slope is negative
⇒ -E_a/R = (10-0)/ (5-0)
E_a = 2R

Without catalyst: $K=A{e}^{-Ea/RT}$

$\begin{array}{rr}& \text{Presence catalyst:}{10}^6\mathrm{K}=A{\mathrm{e}}^{-\mathrm{E}\mathrm{c}/\mathrm{R}\mathrm{T}}\\ & \mathrm{E}\mathrm{q}\left(\mathrm{B}\right)-\mathrm{E}\mathrm{q}\left(\mathrm{A}\right)\mathrm{}{10}^6={\mathrm{e}}^{-(\mathrm{E}-\mathrm{E}\mathrm{c})/\mathrm{R}\mathrm{T}}\\ & \frac{\mathrm{E}-\mathrm{E}\mathrm{c}}{\mathrm{R}\mathrm{T}}=2.303*6\\ & \mathrm{\Delta }\mathrm{E}=\mathrm{E}\mathrm{c}-\mathrm{E}=(-2.303)*6\mathrm{R}\mathrm{T}\end{array}$

Al equilibrium

${\mathrm{K}}_{\mathrm{f}}\left. [{\mathrm{H}}_2\right][\mathrm{N}\mathrm{O}{]}^2=\frac{{\mathrm{K}}_{\mathrm{b}}\left. [{\mathrm{N}}_2\right]{\left. [{\mathrm{H}}_2\mathrm{O}\right]}^{+}}{\left. [{\mathrm{H}}_2\right]}$