Law of Chemical Equilibrium and Equilibrium Constants

 An expression for the backward rate constant of the reaction, $\mathrm{aA}+\mathrm{bB}\stackrel{ }{\rightleftharpoons }\mathrm{cC}+\mathrm{dD}$ is

The reaction $\mathrm{A} \left(\mathrm{g}\right)+\mathrm{B} \left(\mathrm{g}\right)\rightleftharpoons \mathrm{C} \left(\mathrm{g}\right)+\mathrm{D} \left(\mathrm{g}\right)$ is elementary $2^{\mathrm{nd}}$ order reaction opposed by elementary secondary order reaction. When started with equimolar amounts of $\mathrm{A}$ and $\mathrm{B},$ at equilibrium, it is found that the concentration of $\mathrm{A}$ is twice that of $\mathrm{C} .$ Specific rate constant for forward reaction is $2\times {10}^{-3}{\mathrm{mol}}^{-1}{\mathrm{Lsec}}^{-1}.$ The specific rate constant for backward reaction is

For the reaction, $2\mathrm{S}{\mathrm{O}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\leftrightharpoons 2\mathrm{S}{\mathrm{O}}_{3\left(\mathrm{g}\right)}$
What is ${\mathrm{K}}_{\mathrm{c}}$ when the equilibrium concentration of $\left[\mathrm{S}{\mathrm{O}}_2\right]=0.60\mathrm{M}, \left[{\mathrm{O}}_2\right]=0.82\mathrm{M}$ and $\left[\mathrm{S}{\mathrm{O}}_3\right]=1.90\mathrm{M}$?

The reaction

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

is in equilibrium in a closed vessel at $298 \mathrm{K}.$ The partial pressure (in atm) of ${\mathrm{CO}}_2\left(\mathrm{g}\right)$ in the reaction vessel is closest to:

[Given: The change in Gibbs energies of formation at $298 \mathrm{K}$ and $1$ bar for

$\mathrm{CaO}\left(\mathrm{s}\right)=-603.501 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${\mathrm{CO}}_2\left(\mathrm{g}\right)=-394.389 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)=-1128.79 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Gas constant $\mathrm{R}=8.314 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$]

Find the number of reaction which will be favoured in backward direction:

$\left(\mathrm{A}\right)\mathrm{EtOH}+\mathrm{KOH}\rightleftharpoons \mathrm{EtOK}+{\mathrm{H}}_2\mathrm{O}$

$\left(\mathrm{B}\right)$

$\left(\mathrm{C}\right)$

$\left(\mathrm{D}\right)$

Find the approximate value of ${\mathrm{K}}_{\mathrm{c}}$ for the given equilibrium reaction, when $2$ moles of ${\mathrm{SO}}_2$ and $1$ mole of ${\mathrm{O}}_2$ are allowed to react in a container of capacity $2 \mathrm{L}$ and the following equilibrium is established:

$2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{SO}}_3\left(\mathrm{g}\right)$, given that the equilibrium mixture requires $200 \mathrm{mL}$ of $0.1 \mathrm{M}$ acidified ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ solution.

When the two reactants $\mathrm{A}$ and $\mathrm{B}$ are mixed to give products $\mathrm{C}$ and $\mathrm{D}$, the reaction quotient $\mathrm{Q}$ at the initial stage of the reaction 

For the given reaction at equilibrium, ${\mathrm{K}}_{\mathrm{P}}=2.0\times {10}^{10}/\mathrm{bar}$ at $450\mathrm{K}$.         

$2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{SO}}_3\left(\mathrm{g}\right)$

 ${\mathrm{K}}_{\mathrm{c}}$ at this temperature is $\mathrm{n}\times {10}^{11}{\mathrm{Lmol}}^{-1}$. Find value of $\mathrm{n}.$ (Round off n up to the one decimal place)

For the following equilibrium, ${\mathrm{K}}_{\mathrm{c}}= 6.3\times {10}^{14}$ at $1000\mathrm{K}$.
$\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{O}}_3\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{NO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is ${\mathrm{K}}_{\mathrm{c}}$ for the reverse reaction in terms of $\mathrm{n}\times {10}^{-15}$. Find value of n. 

Enter your answer from these options : $(0.5/1.587/1.287/1.785)$

Equilibrium constant for reaction ${\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{NH}}_4^{+}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ is $1.8 \times {10}^9$. Hence equilibrium constant for ionization $\mathrm{N}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{N}{\mathrm{H}}_4^{+}\left(\mathrm{a}\mathrm{q}\right)+\mathrm{O}{\mathrm{H}}^{-}\left(\mathrm{a}\mathrm{q}\right) \mathrm{is} \mathrm{x}\times 10^{-6}$.The value of $\text{'}\mathrm{x}\text{'}$ is:

For the following gaseous equilibrium, ${\mathrm{N}}_2{{\mathrm{O}}_4}_{\left(\mathrm{g}\right)}\rightleftharpoons 2\mathrm{N}{{\mathrm{O}}_2}_{\left(\mathrm{g}\right)}$

${\mathrm{K}}_{\mathrm{p}}$ is found to be equal to ${\mathrm{K}}_{\mathrm{c}}$. This is attained when the temperature is:

For the reaction, A(s) + 2B(g) $\leftrightharpoons$ 2C(g) + D($\mathrm{l}$) at TK
Find the value of K_p when the equilibrium contains 25% of B(g) by volume of the gaseous species.

For the reaction, $2\mathrm{S}{\mathrm{O}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\leftrightharpoons 2\mathrm{S}{\mathrm{O}}_{3\left(\mathrm{g}\right)}$
What is ${\mathrm{K}}_{\mathrm{c}}$ when the equilibrium concentration of $\left[\mathrm{S}{\mathrm{O}}_2\right]=0.60\mathrm{M}, \left[{\mathrm{O}}_2\right]=0.82\mathrm{M}$ and $\left[\mathrm{S}{\mathrm{O}}_3\right]=1.90\mathrm{M}$?

The decay profiles of three radioactive species $A, B$ and $C$ are given below:



These profiles imply that the decay constants $k_A, k_B$ and $k_C$ follow the order

The concentrations of the reactant A in the reaction $A\to B$ at different times are given below

Nine volumes of a gaseous mixture consisting of gaseous organic compound A and just sufficient amount of oxygen required for complete combustion yielded on burning four volumes of $\mathrm{C}{\mathrm{O}}_2$, six volumes of water vapour and two volumes of ${\mathrm{N}}_2$, all volumes measured at the same temperature and pressure. If the compound contains C, H and N only, the molecular formula of the compound A is:

In a chemical reaction, the rate constant for the backward reaction is 7.5 × 10^–4 and the equilibrium constant is 1.5. The rate constant for the forward reaction is -

The rate constant at $25°\mathrm{C}$ for the reaction of ${\mathrm{N}\mathrm{H}}_4^{+}$ & ${\mathrm{O}\mathrm{H}}^{–}$ to form ${\mathrm{N}\mathrm{H}}_4\mathrm{O}\mathrm{H}$ is $4\mathrm{ }\times \mathrm{ }{10}^{10}\mathrm{ }{\mathrm{M}}^{–1}{\mathrm{s}\mathrm{e}\mathrm{c}}^{-1}$& ionisation constant of $\mathrm{a}\mathrm{q}.$ ${\mathrm{N}\mathrm{H}}_3$ is $1.8\mathrm{ }\times \mathrm{ }{10}^{–5}$. The rate constant of proton transfer to ${\mathrm{N}\mathrm{H}}_3$ is -

In a chemical reaction, the rate constant for the backward reaction is $7.5 \times {10}^{–4}$ and the equilibrium constant is $1.5$. The rate constant for the forward reaction is -

In reversible reaction $\mathrm{A}\mathrm{ }+\mathrm{ }\mathrm{B}\rightleftharpoons \mathrm{C}\mathrm{ }+\mathrm{ }\mathrm{D}$, rate constant of forward reaction is ${10}^{–5}$ & rate constant of backward reaction is ${10}^{–3}$. Equilibrium constant $\left({\mathrm{K}}_{\mathrm{C}}\right)$ is -