Write euilibrium constant for the each : (a) N2 O4( g) ⇌2 NO2( g) (b) KClO3( s) ⇌KCl(s)+(3 /

Write euilibrium constant for the each : (a) N2 O4( g) ⇌2...

Write euilibrium constant for the each : (a) $\quad \mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(\mathrm{~g})}$ (b) $\quad \mathrm{KClO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{KCl}_{(\mathrm{s})}+(3 / 2) \mathrm{O}_{2(\mathrm{~g})}$ (c) $\mathrm{CaC}_{2(\mathrm{~s})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{CaCO}_{3(\mathrm{~s})}+2 \mathrm{CO}_{2(\mathrm{~g})}$ (d) $\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}$ (e)Fe $^{3+}{ }_{\text {(aq })}+\mathrm{SCN}_{\text {(aq.) }}^{*}=\mathrm{Fe}(\mathrm{SCN})^{2+}{ }_{\text {(aq })}$ (f) $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} \rightleftharpoons \mathrm{CuSO}_{4(\mathrm{~s})}+5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{v})}$

Write euilibrium constant for the each :
(a) $\quad \mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(\mathrm{~g})}$
(b) $\quad \mathrm{KClO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{KCl}_{(\mathrm{s})}+(3 / 2) \mathrm{O}_{2(\mathrm{~g})}$
(c) $\mathrm{CaC}_{2(\mathrm{~s})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{CaCO}_{3(\mathrm{~s})}+2 \mathrm{CO}_{2(\mathrm{~g})}$
(d) $\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}$
(e)Fe $^{3+}{ }_{\text {(aq })}+\mathrm{SCN}_{\text {(aq.) }}^{*}=\mathrm{Fe}(\mathrm{SCN})^{2+}{ }_{\text {(aq })}$
(f) $\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} \rightleftharpoons \mathrm{CuSO}_{4(\mathrm{~s})}+5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{v})}$

Key Concepts

Law of Mass Action

The law of mass action states that at equilibrium, the ratio of the products' concentrations (or partial pressures for gases) raised to their stoichiometric coefficients to those of the reactants raised to their corresponding coefficients is constant. This ratio is defined as the equilibrium constant (K), which is used to quantify the position of equilibrium for a chemical reaction.

Heterogeneous Equilibria

In heterogeneous equilibria, where reactants and/or products are in different phases (such as gases interacting with solids), only the concentrations (or pressures) of species in the gas or aqueous phase are included in the equilibrium expression. Pure solids and liquids are omitted because their concentrations remain constant, simplifying the equilibrium constant expression.

Chemical Equilibrium

Chemical equilibrium refers to the state in which the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant concentrations of both reactants and products. This dynamic balance is essential for predicting how a reaction behaves under given conditions and forms the basis for calculating equilibrium constants.

Transcript

00:01 While writing equilibrium constant for any equilibrium reaction, keep in mind that the concentration of product is taken on the upper side and concentration of reactants are taken on the lower side in the denominator.

00:25 Also suppose you have any reaction like this small a to capital a plus small b capital b gives so this is the equilibrium sign here c c plus small d capital d so now how will you write this equilibrium constant as i said concentration c into multiplied by concentration d divided by concentration of reactant a multiplied by concentration of reactant b.

01:15 So the concentration of the products are taken on the numerator side on the upper side and the concentration of the reactants are taken in the denominator side and now the stochymetric coefficient returns with them are taken as power.

01:31 So here it will be written as small this d, this is a and this is a.

01:37 B so this is how you will write the equilibrium constant now one thing more to note here if in case of the solid phase reactants or product if your reactant or product is in solid phase then its concentration then its concentration will be taken as unity.

02:46 That is one molar.

02:51 One molar.

02:53 So if you have your reactant or product in solid phase, if you have your reactant or product in solid phase, then the concentration of such a reactant or product in solid phase will be taken as one molar that is unity now we will start our question first part a part n2 o4 gas equilibrium 2 and no 2 gas so the left one is react and the right one is product so it will be kc is equal to n o2 divided by n to 04.

03:59 The concentration are taken in square brackets always.

04:05 Here we have stoichiometric coefficient 2 so it will be written as square to it.

04:13 B part.

04:17 Kl k c l -o -3 solid, equilibrium a cl solid plus 3 upon 2 c 02.

04:40 Gas...

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