K1 = 4.2 x 10–7 and K2 = 4.8 x 10–11
Select the correct statement for a saturated 0.034 M solution of the carbonic acid.
a. CO (g) + H2O (g) $\leftrightharpoons$ CO2(g) + H2 (g) ; K1
b. CH4 (g) + H2O (g) $\leftrightharpoons$ CO(g) + 3H2 (g) ; K2
c. CH4 (g) + 2H2O (g) $\leftrightharpoons$ CO2(g) + 4H2 (g) ; K3
SO3 (g) $\leftrightharpoons$ SO2 (g) + $1 \over 2$ O2 (g)
is Kc = 4.9 $\times$ 10–2. The value of Kc for the reaction
2SO2 (g) + O2 (g) $\leftrightharpoons$ 2SO3 (g) will be :
PCl5 (g) $\leftrightharpoons$ PCl3 (g) + Cl2 (g)
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be
When Kp and Kc are compared at 184oC , it is found that :
Cl2 (g) + 3F2 (g) $\leftrightharpoons$ 2ClF3 (g); $\Delta H$ = -329 kJ
Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3?
P4 (s) + 5O2 $\leftrightharpoons$ P4O10 (s)?
N2O4 (g) $\leftrightharpoons$ 2NO2 (g)
the concentrations of N2O4 and NO2 at equilibrium are 4.8 $\times$ 10-2 and 1.2 $\times$ 10-2 mol L-1 respectively. The value of Kc for the reaction is
2 SO2 (g) + O2 (g) $\leftrightharpoons$ 2 SO3 (g); $\Delta H^o$ = -198 kJ
One the basis of Le Chatelier's principle, the condition favourable for the forward reaction is :
$ \mathrm{X}_2(\mathrm{~g})+\mathrm{Y}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{Z}(\mathrm{~g}) $
$\mathrm{X}_2(\mathrm{~g})$ and $\mathrm{Y}_2(\mathrm{~g})$ are added to a 1 L flask and it is found that the system attains the above equilibrium at $\mathrm{T}(\mathrm{K})$ with the number of moles of $\mathrm{X}_2(\mathrm{~g}), \mathrm{Y}_2(\mathrm{~g})$ and $\mathrm{Z}(\mathrm{g})$ being 3,3 and 9 mol respectively (equilibrium moles). Under this condition of equilibrium, 10 mol of $\mathrm{Z}(\mathrm{g})$ is added to the flask and the temperature is maintained at $\mathrm{T}(\mathrm{K})$. Then the number of moles of $\mathrm{Z}(\mathrm{g})$ in the flask when the new equilibrium is established is $\_\_\_\_$ . (Nearest integer)
Explanation:
Given reaction:
$\mathrm{X_2(g) + Y_2(g) \rightleftharpoons 2Z(g)}$
1) Find $K_c$ from the first equilibrium
At equilibrium in a $1\ \mathrm{L}$ flask:
$ [\mathrm{X_2}] = 3,\quad [\mathrm{Y_2}] = 3,\quad [\mathrm{Z}] = 9 $
$ K_c=\frac{[\mathrm{Z}]^2}{[\mathrm{X_2}][\mathrm{Y_2}]} =\frac{9^2}{3\times 3}=\frac{81}{9}=9 $
2) After adding $10$ mol of $Z$
New moles immediately (before shift):
$ \mathrm{X_2}=3,\quad \mathrm{Y_2}=3,\quad \mathrm{Z}=9+10=19 $
Since $Z$ is added, equilibrium shifts left. Let the backward reaction proceed by $x$ moles:
$\mathrm{X_2}$ increases by $x$
$\mathrm{Y_2}$ increases by $x$
$\mathrm{Z}$ decreases by $2x$
So at new equilibrium:
$ [\mathrm{X_2}] = 3+x,\quad [\mathrm{Y_2}] = 3+x,\quad [\mathrm{Z}] = 19-2x $
3) Apply equilibrium constant again
$ K_c=\frac{(19-2x)^2}{(3+x)(3+x)}=9 $
$ (19-2x)^2=9(3+x)^2 $
Taking positive square root (concentrations are positive):
$ 19-2x = 3(3+x)=9+3x $
$ 19-9 = 3x+2x $
$ 10=5x \Rightarrow x=2 $
Hence,
$ \text{New moles of } Z = 19-2(2)=19-4=15 $
Nearest integer = $15$.
For the following gas phase equilibrium reaction at constant temperature,
$ \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons 1 / 2 \mathrm{~N}_2(\mathrm{~g})+3 / 2 \mathrm{H}_2(\mathrm{~g}) $
if the total pressure is $\sqrt{3} \mathrm{~atm}$ and the pressure equilibrium constant $\left(K_p\right)$ is 9 atm , then the degree of dissociation is given as $\left(x \times 10^{-2}\right)^{-1 / 2}$. The value of $x$ is $\_\_\_\_$ . (nearest integer)
Explanation:
$\mathrm{k}_{\mathrm{P}}=\frac{\left(\frac{\alpha}{2}\right)^{1 / 2}\left(\frac{3 \alpha}{2}\right)^{3 / 2}}{(1-\alpha)}\left[\frac{\mathrm{P}_{\mathrm{T}}}{1+\alpha}\right]^1 \quad\left[\because \mathrm{P}_{\mathrm{T}}=\sqrt{3} \mathrm{~atm}\right]$
$\Rightarrow $ $9=\frac{\left(\frac{\alpha}{2}\right)^{1 / 2}\left(\frac{3 \alpha}{2}\right)^{3 / 2}}{(1-\alpha)} \times \frac{(3)^{1 / 2}}{1+\alpha}$
$\Rightarrow $ $9=\frac{9\left(\frac{\alpha}{2}\right)^2}{1-\alpha^2}$
$\Rightarrow $ $1-\alpha^2=\frac{\alpha^2}{4}$
$\Rightarrow $ $\frac{5 \alpha^2}{4}=1$
$\Rightarrow $ $\alpha^2=0.8$
$\Rightarrow $ $\alpha=(0.8)^{1 / 2}$
$\Rightarrow $ $\alpha=\left[\frac{1}{0.8}\right]^{-1 / 2}$
$\begin{aligned} & \alpha=\left[125 \times 10^{-2}\right]^{-1 / 2} \\\\ &\therefore x=125 .\end{aligned}$
Dissociation of a gas $\mathrm{A}_2$ takes place according to the following chemical reaction. At equilibrium, the total pressure is 1 bar at 300 K .
$ \mathrm{A}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g}) $
The standard Gibbs energy of formation of the involved substances has been provided below:
$ \begin{array}{|c|c|} \hline \text { Substance } & \Delta \mathrm{G}_{\mathrm{f}}^{\circ} / \mathrm{kJ} \mathrm{~mol}^{-1} \\ \hline \hline \mathrm{~A}_2 & -100.00 \\ \hline \mathrm{~A} & -50.832 \\ \hline \end{array} $
The degree of dissociation of $\mathrm{A}_2(\mathrm{~g})$ is given by $\left(x \times 10^{-2}\right)^{1 / 2}$ where $x=$
$\_\_\_\_$ . (Nearest integer).
[ Given : $\mathrm{R}=8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 2=0.3010, \log 3=0.48$ ]
Assume degree of dissociation is not negligible.
Explanation:
$\begin{aligned} & -1.664 \times 10^3=-8.3 \times 300 \ln \mathrm{~K}_{\mathrm{p}} \\ & \ln \mathrm{K}_{\mathrm{p}}=0.693 \\ & \mathrm{~K}_{\mathrm{p}}=2 \\ & 2=\frac{4 \alpha^2 \mathrm{P}_0}{1-\alpha^2} \\ & \alpha=\frac{1}{\sqrt{3}} \\ & \alpha=\left(\frac{100}{3} \times 10^{-2}\right)^{1 / 2} \\ & =\left(33.33 \times 10^{-2}\right)^{1 / 2}\end{aligned}$
The equilibrium constant for decomposition of $\text{H}_2\text{O(g)}$
$ \text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $
is $8.0 \times 10^{-3}$ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($\alpha$) of water is _________ $\times 10^{-2}$ (nearest integer value).
[Assume $\alpha$ is negligible with respect to 1]
Explanation:
$\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}$
$\begin{aligned} & \mathrm{n}_{\mathrm{T}}=1+\frac{\alpha}{2} \simeq 1(\alpha \ll 1) \\ & \mathrm{k}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{H}_2} \cdot \mathrm{P}_{\mathrm{O}_2}^{1 / 2}}{\mathrm{P}_{\mathrm{H}_2 \mathrm{O}}}=\frac{(\alpha \cdot \mathrm{P})\left(\frac{\alpha}{2} \mathrm{P}\right)^{\frac{1}{2}}}{(1-\alpha) \mathrm{P}} \\ & \mathrm{P}=1 \\ & 8 \times 10^{-3}=\frac{\alpha^{3 / 2}}{\sqrt{2}} \\ & \alpha^{3 / 2}=8 \sqrt{2} \times 10^{-3} \\ & \alpha^3=128 \times 10^{-6} \\ & \alpha=\sqrt[3]{128} \times 10^{-2} \\ & =5.03 \times 10^{-2} \end{aligned}$
Consider the following equilibrium,
$\mathrm{CO}(\mathrm{~g})+2 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{~g})$
0.1 mol of CO along with a catalyst is present in a $2 \mathrm{dm}^3$ flask maintained at 500 K . Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of $\mathrm{CH}_3 \mathrm{OH}$ is formed. The $K_p^\theta$ is __________ $\times 10^{-3}$ (nearest integer).
Given : $\mathrm{R}=0.08 \mathrm{~dm}^3$ bar $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$
Assume only methanol is formed as the product and the system follows ideal gas behaviour.
Explanation:
$\begin{aligned} & V=2 L \\ & T=500 K \\ & P_{\text {total }}=5 \text { bar } \\ & n_{\text {Total }}=0.25=\frac{1}{4} \mathrm{~mol} \\ & P_{\text {total }}=n_{\text {total }} \times \frac{R T}{V} \\ & \Rightarrow 5=(0.06+a-0.08+0.04) \times \frac{0.08 \times 500}{2} \\ & \Rightarrow 10=(0.02+a) \times 0.08 \times 500 \end{aligned}$
$\begin{aligned} &\Rightarrow \mathrm{a}=0.25-0.02=0.23 \mathrm{~mol}\\ &\begin{aligned} & \mathrm{K}_{\mathrm{P}}=\frac{\mathrm{X}_{\mathrm{CH}_3 \mathrm{OH}}}{\mathrm{X}_{\mathrm{CO}} \times \mathrm{X}_{\mathrm{H}_2}^2} \times \frac{1}{\left(\mathrm{P}_{\mathrm{T}}\right)^2}=\frac{0.04}{0.06 \times(0.15)^2} \times\left[\frac{1 / 4}{5}\right]^2 \\ & =\frac{4}{6 \times(0.15)^2 \times 16} \times \frac{1}{25} \\ & =\frac{100 \times 100}{24 \times 225 \times 25}=\frac{100 \times 100}{135000} \\ & =0.074=74 \times 10^{-3} \end{aligned} \end{aligned}$
$37.8 \mathrm{~g} \mathrm{~N}_2 \mathrm{O}_5$ was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K
$2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_{4(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
The total pressure at equilibrium was found to be 18.65 bar.
Then, $\mathrm{Kp}=$ _________ $\times 10^{-2}$ [nearest integer]
Assume $\mathrm{N}_2 \mathrm{O}_5$ to behave ideally under these conditions.
Given: $\mathrm{R}=0.082$ bar $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~K}^{-1}$
Explanation:
$\begin{aligned} & \text { Initial pressure of } \mathrm{N}_2 \mathrm{O}_5 \\ & \qquad \\ & \qquad=\frac{\frac{37.8}{108} \times 0.082 \times 500}{1}=14.35 \mathrm{bar} \\ & 2 \mathrm{~N}_2 \mathrm{O}_5 \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_4+\mathrm{O}_2 \end{aligned}$
$\begin{aligned} & \mathrm{t}=0 \quad 14.35 \\ & \mathrm{t}=\mathrm{eq} \quad 14.35-2 \mathrm{P} \quad 2 \mathrm{P} \quad \mathrm{P} \\ & \mathrm{P}_{\text {Total }} \text { at eqb }=14.35+\mathrm{P}=18.65 \\ & \mathrm{P}=4.3 \\ & \mathrm{P}_{\mathrm{N}_2 \mathrm{O}_{\mathrm{s}}}=5.75 \mathrm{bar} \\ & \mathrm{P}_{\mathrm{N}_2 \mathrm{O}_4}=8.6 \mathrm{bar} \\ & \mathrm{P}_{\mathrm{O}_2}=4.3 \mathrm{bar} \\ & \mathrm{k}_{\mathrm{p}}=\frac{(8.6)^2 \times(4.3)}{(5.75)^2}=9.619=\mathrm{x} \times 10^{-2} \\ & \mathrm{x}=961.9 \approx 962 \end{aligned}$
The following concentrations were observed at $500 \mathrm{~K}$ for the formation of $\mathrm{NH}_3$ from $\mathrm{N}_2$ and $\mathrm{H}_2$. At equilibrium ; $\left[\mathrm{N}_2\right]=2 \times 10^{-2} \mathrm{M},\left[\mathrm{H}_2\right]=3 \times 10^{-2} \mathrm{M}$ and $\left[\mathrm{NH}_3\right]=1.5 \times 10^{-2} \mathrm{M}$. Equilibrium constant for the reaction is ________.
Explanation:
To calculate the equilibrium constant $\mathrm{K}_C$ for the formation of $\mathrm{NH}_3$ from $\mathrm{N}_2$ and $\mathrm{H}_2$ at $500 \mathrm{~K}$, we use the formula:
$ \mathrm{K}_C = \frac{[\mathrm{NH}_3]^2}{[\mathrm{N}_2][\mathrm{H}_2]^3} $
Substituting the given concentrations:
$ \mathrm{K}_C = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2}) \times (3 \times 10^{-2})^3} $
Calculating the above expression results in:
$ \mathrm{K}_C = 417 $
For the reaction $\mathrm{N}_2 \mathrm{O}_{4(\mathrm{~g})} \rightleftarrows 2 \mathrm{NO}_{2(\mathrm{~g})}, \mathrm{K}_{\mathrm{p}}=0.492 \mathrm{~atm}$ at $300 \mathrm{~K} . \mathrm{K}_{\mathrm{c}}$ for the reaction at same temperature is _________ $\times 10^{-2}$.
(Given : $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$)
Explanation:
For the reaction:
$ \mathrm{N}_2 \mathrm{O}_{4(\mathrm{~g})} \rightleftarrows 2 \mathrm{NO}_{2(\mathrm{~g})} $
we need to find the equilibrium constant $\mathrm{K}_{\mathrm{c}}$ at 300 K given that $\mathrm{K}_{\mathrm{p}} = 0.492 \text{ atm}$.
The relationship between $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{c}}$ is given by the following equation:
$ \mathrm{K}_{\mathrm{p}} = \mathrm{K}_{\mathrm{c}} (RT)^{\Delta n} $
where:
$R = 0.082 \, \mathrm{L} \, \mathrm{atm} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$ is the gas constant,
$T = 300 \, \mathrm{K}$ is the temperature,
$\Delta n$ is the change in moles of gas from reactants to products.
For the reaction given:
The number of moles of gaseous products is 2 (for $2 \mathrm{NO}_2$).
The number of moles of gaseous reactants is 1 (for $\mathrm{N}_2 \mathrm{O}_4$).
Thus, $\Delta n = 2 - 1 = 1$.
Substitute the known values into the equation:
$ 0.492 = \mathrm{K}_{\mathrm{c}} (0.082 \times 300)^{1} $
Solve for $\mathrm{K}_{\mathrm{c}}$:
$ 0.492 = \mathrm{K}_{\mathrm{c}} \times 24.6 $
$ \mathrm{K}_{\mathrm{c}} = \frac{0.492}{24.6} $
Calculate $\mathrm{K}_{\mathrm{c}}$:
$ \mathrm{K}_{\mathrm{c}} = 0.02 $
Thus, $\mathrm{K}_{\mathrm{c}}$ for the reaction at 300 K is:
$ 2 \times 10^{-2} $
4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of $\mathrm{HI}$ were found. The equilibrium constant for $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ is _________.
Explanation:
$ \mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{(3)^2}{3 \times 3}=\frac{9}{9}=1 $
A mixture of 1 mole of $\mathrm{H}_{2} \mathrm{O}$ and 1 mole of $\mathrm{CO}$ is taken in a 10 litre container and heated to $725 \mathrm{~K}$. At equilibrium $40 \%$ of water by mass reacts with carbon monoxide according to the equation :
$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})$.
The equilibrium constant $\mathrm{K}_{\mathrm{c}} \times 10^{2}$ for the reaction is ____________. (Nearest integer)
Explanation:
$\mathrm{CO(g)} + \mathrm{H_{2}O(g)} \rightleftharpoons \mathrm{CO_{2}(g)} + \mathrm{H_{2}(g)} $
We are given that $40\%$ of water by mass reacts with carbon monoxide.
The molecular weight of water (H2O) is approximately $18 \, \mathrm{g/mol}$, so $1 \, \mathrm{mole}$ of water weighs $18 \, \mathrm{g}$. Therefore, the mass of the reacted water is $0.40 \times 18 \, \mathrm{g} = 7.2 \, \mathrm{g}$.
Since from the stoichiometry of the reaction we can see that $1 \, \mathrm{mole}$ of CO reacts with $1 \, \mathrm{mole}$ of H2O to form $1 \, \mathrm{mole}$ of CO2 and $1 \, \mathrm{mole}$ of H2, this means that $7.2 \, \mathrm{g}$ of H2O is equivalent to $7.2 \, \mathrm{g} / 18 \, \mathrm{g/mol} = 0.4 \, \mathrm{mole}$.
We start with $1 \, \mathrm{mole}$ of CO and $1 \, \mathrm{mole}$ of H2O. At equilibrium, we have :
- CO: $1 \, \mathrm{mole} - 0.4 \, \mathrm{mole} = 0.6 \, \mathrm{mole}$
- H2O : $1 \, \mathrm{mole} - 0.4 \, \mathrm{mole} = 0.6 \, \mathrm{mole}$
- CO2 : $0 \, \mathrm{mole} + 0.4 \, \mathrm{mole} = 0.4 \, \mathrm{mole}$
- H2 : $0 \, \mathrm{mole} + 0.4 \, \mathrm{mole} = 0.4 \, \mathrm{mole}$
The volume of the container is $10 \, \mathrm{litres}$. Therefore, we can convert the moles to concentrations (in M = moles/L) as :
- [CO] = $0.6 \, \mathrm{M}$
- [H2O] = $0.6 \, \mathrm{M}$
- [CO2] = $0.4 \, \mathrm{M}$
- [H2] = $0.4 \, \mathrm{M}$
The equilibrium constant $K_{c}$ for the reaction can be calculated as :
$K_{c} = \frac{[\mathrm{CO_{2}}][\mathrm{H_{2}}]}{[\mathrm{CO}][\mathrm{H_{2}O}]}$
So, substituting the values we get :
$K_{c} = \frac{(0.4 \times 0.4)}{(0.6 \times 0.6)} = 0.44$
As per the question, we need to calculate the value of $K_{c} \times 10^{2}$ :
$K_{c} \times 10^{2} = 0.44 \times 10^{2} = 44 \, (\text{rounded to the nearest integer})$
Therefore, $K_{c} \times 10^{2} = 44$.
$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)+\mathrm{C}(g)$
For the given reaction, if the initial pressure is $450 \mathrm{~mm} ~\mathrm{Hg}$ and the pressure at time $\mathrm{t}$ is $720 \mathrm{~mm} ~\mathrm{Hg}$ at a constant temperature $\mathrm{T}$ and constant volume $\mathrm{V}$. The fraction of $\mathrm{A}(\mathrm{g})$ decomposed under these conditions is $x \times 10^{-1}$. The value of $x$ is ___________ (nearest integer)
Explanation:
Given the reaction:
$\mathrm{A}_{(\mathrm{g})} \rightleftharpoons 2 \mathrm{~B}_{(\mathrm{g})}+\mathrm{C}_{(\mathrm{g})}$
At the beginning of the reaction (at time = 0), the total pressure is solely due to A, hence it is 450 mmHg.
As the reaction progresses, let's denote 'x' as the pressure decrease due to the decomposition of A. Correspondingly, the pressure increases by '2x' and 'x' for B and C respectively, following the stoichiometry of the reaction.
At time 't', the total pressure (P(T)) is the sum of the pressures of A, B, and C, which is given to be 720 mmHg.
This gives us the equation:
$450 \text{ mmHg} - x \text{ mmHg (decrease in A's pressure)} + 2x \text{ mmHg (increase in B's pressure)} + x \text{ mmHg (increase in C's pressure)} = 720 \text{ mmHg (total pressure at time t)}$
Solving for 'x' gives:
$x = 135 \text{ mmHg}$
The fraction of A decomposed would then be this change in pressure divided by the initial pressure:
$\text{Fraction decomposed} = \frac{x}{\text{Initial pressure}} = \frac{135 \text{ mmHg}}{450 \text{ mmHg}} = 0.3 = 3 \times 10^{-1}$
The number of correct statement/s involving equilibria in physical processes from the following is ________
(A) Equilibrium is possible only in a closed system at a given temperature.
(B) Both the opposing processes occur at the same rate.
(C) When equilibrium is attained at a given temperature, the value of all its parameters became equal.
(D) For dissolution of solids in liquids, the solubility is constant at a given temperature.
Explanation:
(A) Equilibrium is possible only in a closed system at a given temperature.
- This statement is true. In a closed system, there is no exchange of matter with the surroundings. This allows the reaction to reach equilibrium at a constant temperature.
(B) Both the opposing processes occur at the same rate.
- This statement is true. At equilibrium, the forward and reverse reactions occur at the same rate, meaning the concentrations of reactants and products remain constant.
(C) When equilibrium is attained at a given temperature, the value of all its parameters became equal.
- This statement is false. When a system reaches equilibrium, it doesn't mean that the values of all parameters (such as the concentrations of reactants and products) are equal. It means that these values are constant, i.e., they are not changing with time because the rates of the forward and reverse reactions are equal.
(D) For dissolution of solids in liquids, the solubility is constant at a given temperature.
- This statement is true. The solubility of a solid in a liquid (in a saturated solution) is indeed constant at a given temperature and pressure.
Therefore, there are 3 correct statements: (A), (B), and (D).
The equilibrium composition for the reaction $\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}$ at $298 \mathrm{~K}$ is given below:
$\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=0.2 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=0.1 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=0.40 \mathrm{~mol} \mathrm{~L}^{-1}$
If $0.2 \mathrm{~mol}$ of $\mathrm{Cl}_{2}$ is added at the same temperature, the equilibrium concentrations of $\mathrm{PCl}_{5}$ is __________ $\times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$
Given : $\mathrm{K}_{\mathrm{c}}$ for the reaction at $298 \mathrm{~K}$ is 20
Explanation:
The initial equilibrium concentrations are given as:
$[\mathrm{PCl}_{3}]_{\text{eq}} = 0.2 \, \mathrm{mol/L}$, $[\mathrm{Cl}_{2}]_{\text{eq}} = 0.1 \, \mathrm{mol/L}$, $[\mathrm{PCl}_{5}]_{\text{eq}} = 0.4 \, \mathrm{mol/L}$.
After adding $0.2 \, \mathrm{mol}$ of $\mathrm{Cl_2}$, the new concentration of $\mathrm{Cl_2}$ becomes $0.1 + 0.2 = 0.3 \, \mathrm{mol/L}$.
Let the change in concentration be $x$, so at the new equilibrium, the concentrations are:
$[\mathrm{PCl}_{3}] = 0.2 - x \, \mathrm{mol/L}$, $[\mathrm{Cl}_{2}] = 0.3 - x \, \mathrm{mol/L}$, $[\mathrm{PCl}_{5}] = 0.4 + x \, \mathrm{mol/L}$.
Using the equilibrium constant expression $K_c = 20$, we get:
$20 = \frac{0.4+x}{(0.2-x)(0.3-x)}$.
Solving for $x$, we find $x \approx 0.086$.
Therefore, the new equilibrium concentration of $\mathrm{PCl}_5$ is:
$[\mathrm{PCl}_5]_{\text{new eq}} = 0.4 + 0.086 = 0.486 \, \mathrm{mol/L} = 48.6 \times 10^{-2} \, \mathrm{mol/L}$.
(i) $\mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 1}=3$
(ii) $\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 2}=1$
If the degree of dissociation and initial concentration of both the reactants $\mathrm{X}(\mathrm{g})$ and $\mathrm{A}(\mathrm{g})$ are equal, then the ratio of the total pressure at equilibrium $\left(\frac{p_{1}}{p_{2}}\right)$ is equal to $\mathrm{x}: 1$. The value of $\mathrm{x}$ is _____________ (Nearest integer)
Explanation:
$ \begin{aligned} & \mathrm{k}_{\mathrm{p}_1}=\frac{\left(\frac{\alpha}{1+\alpha} \times \mathrm{p}_1\right)^2}{\frac{1-\alpha}{1+\alpha} \mathrm{p}_1} \\\\ & 3=\frac{\alpha^2 \times \mathrm{p}_1}{1-\alpha^2} \end{aligned} $
$ \begin{aligned} & \mathrm{k}_{\mathrm{p}_2}=\frac{\left(\frac{2 \alpha}{1+\alpha} \times \mathrm{p}_2\right)^2}{\frac{1-\alpha}{1+\alpha} \times \mathrm{p}_2} \\\\ & 1=\frac{4 \alpha^2 \times \mathrm{p}_2}{1-\alpha^2} \\\\ & \frac{\mathrm{k}_{\mathrm{p}_1}}{\mathrm{k}_{\mathrm{p}_2}}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\ & \frac{3}{1}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\ & {\mathrm{p}_1} : {\mathrm{p}_2} = 12 : 1 \\\\ & \therefore \mathrm{x}=12 \end{aligned} $
For reaction : $\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})$
$\mathrm{K}_{\mathrm{p}}=2 \times 10^{12}$ at $27^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure. The $\mathrm{K}_{\mathrm{c}}$ for the same reaction is ____________ $\times 10^{13}$. (Nearest integer)
(Given $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$)
Explanation:
$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), \Delta H=-190 \mathrm{~kJ}$
The number of factors which will increase the yield of $\mathrm{SO}_{3}$ at equilibrium from the following is _______.
A. Increasing temperature
B. Increasing pressure
C. Adding more $\mathrm{SO}_{2}$
D. Adding more $\mathrm{O}_{2}$
E. Addition of catalyst
Explanation:
$\mathrm{2SO_2+O_2\rightleftharpoons 2SO_3~~\Delta H=-190~kJ}$
It is an exothermic reaction
$\therefore$ factor B, C, D will increase the amount of SO$_3$.
At 298 K
$\mathrm{N_2~(g)+3H_2~(g)\rightleftharpoons~2NH_3~(g),~K_1=4\times10^5}$
$\mathrm{N_2~(g)+O_2~(g)\rightleftharpoons~2NO~(g),~K_2=1.6\times10^{12}}$
$\mathrm{H_2~(g)+\frac{1}{2}O_2~(g)\rightleftharpoons~H_2O~(g),~K_3=1.0\times10^{-13}}$
Based on above equilibria, then equilibrium constant of the reaction, $\mathrm{2NH_3(g)+\frac{5}{2}O_2~(g)\rightleftharpoons~2NO~(g)+3H_2O~(g)}$ is ____________ $\times10^{-33}$ (Nearest integer).
Explanation:
$\mathrm{2NH_3(g)\frac{5}{2}O_2(g)\rightleftharpoons 2NO(g)+3H_2O(g)}$
Clearly, $\mathrm{{K_{eq}} = {1 \over {{k_1}}} \times {k_2} \times k_3^3}$
$ = {{1.6 \times {{10}^{12}} \times {{10}^{ - 39}}} \over {4 \times {{10}^5}}}$
$ = 0.4 \times {10^{ - 32}}$
$ = 4 \times {10^{ - 33}}$
Water decomposes at 2300 K
$\mathrm{H_2O(g)\to H_2(g)+\frac{1}{2}O_2(g)}$
The percent of water decomposing at 2300 K and 1 bar is ___________ (Nearest integer).
Equilibrium constant for the reaction is $2\times10^{-3}$ at 2300 K.
Explanation:
$ \begin{aligned} & \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{T}}^{\frac{3}{2}}\left(1+\frac{\alpha}{2}\right) \alpha^{\frac{3}{2}}}{2^{\frac{1}{2}} \mathrm{P}_{\mathrm{T}}\left(1+\frac{\alpha}{2}\right)^{\frac{3}{2}}(1-\alpha)} \\\\ & 2 \times 10^{-3}=\frac{\alpha^{\frac{3}{2}}}{2^{\frac{1}{2}}}\left[\text { as } \alpha<<1 \text { and let } \mathrm{P}_{\mathrm{T}}=1\right] \end{aligned} $
$ \begin{aligned} & \alpha^{\frac{3}{2}}=2^{\frac{3}{2}} \times 10^{-3} \\\\ & \approx 2 \times 10^{-2} \end{aligned} $
$\therefore \%$ of water decomposition $=2 \%$
Consider the following reaction approaching equilibrium at 27$^\circ$C and 1 atm pressure
$\mathrm{A+B}$ $\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{{k_r} = {{10}^2}}^{{k_f} = {{10}^3}}} $ $\mathrm{C+D}$
The standard Gibb's energy change $\mathrm{(\Delta_r G^\theta)}$ at 27$^\circ$C is ($-$) ___________ kJ mol$^{-1}$ (Nearest integer).
(Given : $\mathrm{R=8.3~J~K^{-1}~mol^{-1}}$ and $\mathrm{\ln 10=2.3}$)
Explanation:
The Gibbs energy change for a reaction at standard conditions, $\Delta_r G^\theta$, can be calculated using the equation:
$ \Delta \mathrm{G}^\theta = -\mathrm{RT} \ln \mathrm{K}_{\mathrm{eq}} $
Where $\mathrm{K}_{\mathrm{eq}}$ is the equilibrium constant given by $\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}$. For the reaction:
$ \mathrm{A + B} \rightleftharpoons \mathrm{C + D} $
The forward rate constant, $k_f$, is $10^3$, and the reverse rate constant, $k_r$, is $10^2$. Therefore,
$ \mathrm{K}_{\mathrm{eq}} = \frac{10^3}{10^2} = 10 $
Substitute the values into the Gibbs energy change formula:
$ \Delta \mathrm{G}^\theta = -\mathrm{RT} \ln 10 $
where $R = 8.3 \, \mathrm{J \, K^{-1} \, mol^{-1}}$ and $T = 300 \, \mathrm{K}$ (since the temperature is $27^\circ \mathrm{C}$ which converts to 300 K). The natural logarithm of 10 is approximately 2.3.
$ \Delta \mathrm{G}^\theta = -(8.3 \times 300 \times 2.3) $
Calculating gives:
$ \Delta \mathrm{G}^\theta = -5711 \, \mathrm{J \, mol^{-1}} $
Converting to kJ:
$ \Delta \mathrm{G}^\theta = -5.7 \, \mathrm{kJ \, mol^{-1}} $
Rounding to the nearest integer, the standard Gibb's energy change is approximately $-6 \, \mathrm{kJ \, mol^{-1}}$.
At $600 \mathrm{~K}, 2 \mathrm{~mol}$ of $\mathrm{NO}$ are mixed with $1 \mathrm{~mol}$ of $\mathrm{O}_{2}$.
$2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g})$
The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that $0.6 \mathrm{~mol}$ of oxygen are present at equilibrium. The equilibrium constant for the reaction is ________. (Nearest integer)
Explanation:
Partial pressure of $\mathrm{NO}(\mathrm{g})=\frac{1.2}{2.6} \times 1$
Partial pressure of $\mathrm{O}_{2}(\mathrm{~g})=\frac{0.6}{2.6}$
Partial pressure of $\mathrm{NO}_{2}(\mathrm{~g})=\frac{0.8}{2.6}$
$ \begin{aligned} \mathrm{K}_{\mathrm{p}}=\frac{\left(\mathrm{P}_{\mathrm{NO}_{2}}\right)^{2}}{\left(\mathrm{P}_{\mathrm{NO}}\right)^{2}\left(\mathrm{P}_{\mathrm{O}_{2}}\right)} &=\frac{0.8 \times 0.8 \times 2.6}{1.2 \times 1.2 \times 0.6} \\\\ &=1.925 \\\\ & \approx 2 \end{aligned} $
At $298 \mathrm{~K}$, the equilibrium constant is $2 \times 10^{15}$ for the reaction :
$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
The equilibrium constant for the reaction
$ \frac{1}{2} \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s}) \rightleftharpoons \frac{1}{2} \mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) $
is $x \times 10^{-8}$. The value of $x$ is _____________. (Nearest Integer)
Explanation:
$k=2 \times 10^{15}$
$\frac{1}{2} \mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{+2}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
$\mathrm{K}^{\prime}=\frac{1}{(\mathrm{~K})^{1 / 2}}=\frac{1}{\left(2 \times 10^{15}\right)^{1 / 2}}$
$=2.23 \times 10^{-8}$
$x \simeq 2$
A box contains 0.90 g of liquid water in equilibrium with water vapour at 27$^\circ$C. The equilibrium vapour pressure of water at 27$^\circ$C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be __________ litre. [nearest integer]
(Given : R = 0.082 L atm K$-$1 mol$-$1)
(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)
Explanation:
We know, 760 Torr = 1 atm
$\therefore$ 32 Torr = ${{32} \over {760}}$ atm
As all the liquid water evaporates so entire water is in gaseous state.
$\therefore$ Weight of water vapour = 0.9 g
$\therefore$ Moles of water vapour (n) = ${{0.9} \over {18}}$
Pressure (P) = ${{32} \over {760}}$ atm
Temperature (T) = (27 + 273) K = 300 K
R = 0.082 L atm K$-$1 mol$-$1
Given water vapour act as an ideal gas, so we can apply ideal gas equation.
From ideal gas equation,
PV = nRT
$ \Rightarrow {{32} \over {760}} \times v = {{0.9} \over {18}} \times 0.082 \times 300$
$ \Rightarrow v = 29$ L
2NOCl(g) $\rightleftharpoons$ 2NO(g) + Cl2(g)
In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30$^\circ$C is ______________ $\times$ 10$-$4.
Explanation:
Given that at equilibrium, concentration of NO = 0.4 mol/L
$\therefore$ 2x = 0.4
$\Rightarrow$ x = 0.2
$\therefore$ Concentration of NOCl at equilibrium,
[NOCl]eq = 2 $-$ 2 $\times$ 0.2 = 1.6
and [NO]eq = 0.4
and [Cl2]eq = 0.2
We know,
${K_C} = {{{{[NO]}^2}[C{l_2}]} \over {{{[NOCl]}^2}}}$
$ = {{{{[0.4]}^2}[0.2]} \over {{{[1.6]}^2}}}$
$ \Rightarrow {K_C} = 12.5 \times {10^{ - 3}}$
$ \Rightarrow {K_C} = 125 \times {10^{ - 4}}$
40% of HI undergoes decomposition to H2 and I2 at 300 K. $\Delta$G$^\Theta $ for this decomposition reaction at one atmosphere pressure is __________ J mol$-$1. [nearest integer]
(Use R = 8.31 J K$-$1 mol$-$1 ; log 2 = 0.3010, ln 10 = 2.3, log 3 = 0.477)
Explanation:
$\therefore$ $K = {{{{\left( {{\alpha \over 2}} \right)}^{1/2}} \times {{\left( {{\alpha \over 2}} \right)}^{1/2}}} \over {(1 - \alpha )}}$
Given $\alpha = {{40} \over {100}} = 0.4$
$\therefore$ $K = {{{{\left( {{{0.4} \over 2}} \right)}^{1/2}} \times {{\left( {{{0.4} \over 2}} \right)}^{1/2}}} \over {(1 - 0.4)}}$
$ = {1 \over 3}$
We know,
$\Delta G^\circ = - RT\ln K$
$ = - RT\ln \left( {{1 \over 3}} \right)$
$ = + RT\ln 3$
$ = + 8.314 \times 300 \times \ln 3$
= 2735 J/mol
The standard free energy change ($\Delta$G$^\circ$) for 50% dissociation of N2O4 into NO2 at 27$^\circ$C and 1 atm pressure is $-$ x J mol$-$1. The value of x is ___________. (Nearest Integer)
[Given : R = 8.31 J K$-$1 mol$-$1, log 1.33 = 0.1239 ln 10 = 2.3]
Explanation:
$\mathrm{k}_{\mathrm{P}}=\frac{\left(\frac{1}{1.5} \times 1\right)^2}{\left(\frac{0.5}{1.5} \times 1\right)}=\frac{1}{0.75}=\frac{100}{75}$
$=1.33$
$\Delta \mathrm{G}^0=-\mathrm{RT} \ell \mathrm{nk}_{\mathrm{P}}$
$=-8.31 \times 300 \times \ln (1.33)=-710.45 \mathrm{~J} / \mathrm{mol}$
$=-710 \mathrm{~J} / \mathrm{mol}$
PCl5 dissociates as
PCl5(g) $\rightleftharpoons$ PCl3(g) + Cl2(g)
5 moles of PCl5 are placed in a 200 litre vessel which contains 2 moles of N2 and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5 is __________ $\times$ 10$-$3. (nearest integer)
(Given : R = 0.082 L atm K$-$1 mol$-$1; Assume ideal gas behaviour)
Explanation:
Here 2 moles of N2 also present that is why 2 moles always have to add in total mole calculation.
At equilibrium,
Pressure (P) = 2.46 atm
Volume (V) = 200 L
Temperature (T) = 600 K
$\therefore$ Applying ideal gas equation,
PV = nRT
$\Rightarrow$ 2.46 $\times$ 200 = (7 + x) $\times$ 0.082 $\times$ 600
$\Rightarrow$ x = 3
Now,
${K_P} = {{{P_{PC{l_3}}} \times {P_{C{l_2}}}} \over {{P_{PC{l_5}}}}}$
$ = {{\left[ {{3 \over {7 + 3}} \times 2.46} \right]\left[ {{3 \over {7 + 3}} \times 2.46} \right]} \over {\left[ {{{5 - 3} \over {7 + 3}} \times 2.46} \right]}}$
$ = {{{3 \over {10}} \times {3 \over {10}} \times {{(2.46)}^2}} \over {{2 \over {10}} \times 2.46}}$
$ = {9 \over {20}} \times 2.46$
$ = 1107 \times {10^{ - 3}}$ atm
2O3(g) $\rightleftharpoons$ 3O2(g)
At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is ($-$) ____________ J mol$-$1. (Nearest integer)
[Given : ln 1.35 = 0.3 and R = 8.3 J K$-$1 mol$-$1]
Explanation:
Given, $x=0.5$
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{[3(0.5)]^{3} \times 1}{[2]^{3} \times(0.5)^{2} \times 1.25}$
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{27}{8} \times \frac{0.5}{1.25}=1.35$
$ \begin{aligned} \Delta \mathrm{G}^{\circ} &=-2.303 \mathrm{RT} \log \mathrm{k}_{\mathrm{p}} \\\\ &=-2.303 \times 8.3 \times 300 \log 1.35 \\\\ &=-8.3 \times 300 \ln (1.35) \\\\ &=-747 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} $
Explanation:
volume of vessel = 2 litre
$\Rightarrow$ partial pressure of each component
$P = {{nRT} \over V} = {{0.1 \times 0.2 \times 0.082 \times 300} \over 2}$
= 0.246 atm
$\Rightarrow$ kP = P$N{H_3}$ $\times$ P${H_2}S$ = (0.246)2 = 0.060516
= 6.05 $\times$ 10$-$2
$ \therefore $ x = 6
[Assume no volume change on adding NH3]
Explanation:
${{0.8} \over {(5 \times {{10}^{ - 8}})\left( {{a \over 2} - 1.6} \right)}} = {10^8}$
$\Rightarrow$ ${{a \over 2}}$ $-$ 1.6 = 0.4 $\Rightarrow$ a = 4
Explanation:
$\therefore$ ${K_C} = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$
$100 = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$
${{1 + x} \over {1 - x}} = 10$
$x = {9 \over {11}}$
Moles of D = 1 + x
$ = 1 + {9 \over {11}} = {{20} \over {11}}$
$ = 1.818 = 181.8 \times {10^{ - 2}} = 181.8 \times {10^{ - 2}}$
$ \cong 182 \times {10^{ - 2}}$ M
[PtCl4]2$-$ + H2O $\rightleftharpoons$ [Pt(H2O)Cl3]$-$ + Cl$-$
was measured as a function of concentrations of different species. It was observed that ${{ - d\left[ {{{\left[ {PtC{l_4}} \right]}^{2 - }}} \right]} \over {dt}} = 4.8 \times {10^{ - 5}}\left[ {{{\left[ {PtC{l_4}} \right]}^{2 - }}} \right] - 2.4 \times {10^{ - 3}}\left[ {{{\left[ {Pt({H_2}O)C{l_3}} \right]}^ - }} \right]\left[ {C{l^ - }} \right]$.
where square brackets are used to denote molar concentrations. The equilibrium constant Kc = ____________ . (Nearest integer)
Explanation:
$k_f[\text{{PtCl}}_4]^{2-} = k_r[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$
Given the rate equation:
$-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$
At equilibrium, $-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 0$, so:
$0 = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$
Rearranging terms, we find:
$4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} = 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$
Now, the equilibrium constant $K_c$ is defined as the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. For the reaction in question, we have:
$K_c = \frac{[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]}{[\text{{PtCl}}_4]^{2-}}$
Dividing both sides of our rate equation by $[\text{{PtCl}}_4]^{2-}$, we find that:
$K_c = \frac{4.8 \times 10^{-5}}{2.4 \times 10^{-3}} = \frac{1}{50}$ = 0.02
So, the equilibrium constant $K_c$ for this reaction is approximately 0, when rounded to the nearest integer.
[Given Kw = 1 $\times$ 10$-$14 and Kb = 1.8 $\times$ 10$-$5]
Explanation:
So, ${K_b} = {{[NH_4^ + ][H{O^ - }]} \over {[N{H_3}]}}$
$[H{O^ - }] = {{{K_b} \times [N{H_3}]} \over {[NH_4^ + ]}} = 1.8 \times {10^{ - 5}} \times {2 \over 5} \times {{210} \over {504}} = 3 \times {10^{ - 6}}$
A(s) $\rightleftharpoons$ M(s) + ${1 \over 2}$O2(g)
is Kp = 4. At equilibrium, the partial pressure of O2 is _________ atm. (Round off to the nearest integer)
Explanation:
In the given equilibrium, the solid substances do not contribute to the equilibrium constant expression since their activities are considered to be 1.
For the reaction:
$ \text{A(s)} \rightleftharpoons \text{M(s)} + \frac{1}{2}\text{O}_2(\text{g}) $
The equilibrium constant, $ K_p $, is given by:
$ K_p = P_{\text{O}_2}^{n} $
where $ P_{\text{O}_2} $ is the partial pressure of $ \text{O}_2 $ and $ n $ represents the stoichiometric coefficient of $ \text{O}_2 $ in the balanced chemical reaction, which is $ \frac{1}{2} $. Therefore, the expression for $ K_p $ becomes:
$ K_p = \left( P_{\text{O}_2} \right)^{\frac{1}{2}} $
Given that $ K_p = 4 $, substituting into the equation gives:
$ 4 = \left( P_{\text{O}_2} \right)^{\frac{1}{2}} $
To find $ P_{\text{O}_2} $, square both sides of the equation:
$ 4^2 = P_{\text{O}_2} $
$ 16 = P_{\text{O}_2} $
Thus, the partial pressure of $ \text{O}_2 $ at equilibrium is $\boxed{16}$ atm.
Kc = 1.844
3.0 moles of PCl5 is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl5 at equilibrium is ______________ $\times$ 10$-$3. (Round off to the Nearest Integer)
Explanation:
t = 0 3moles
t = $\infty$ x x
$ \Rightarrow {{[PC{l_3}][C{l_2}]} \over {[PC{l_5}]}} = {{{x^2}} \over {3 - x}} = 1.844$
$ \Rightarrow {x^2} + 1.844 - 5.532 = 0$
$ \Rightarrow x = {{ - 1.844 + \sqrt {{{(1.844)}^2} + 4 \times 5.532} } \over 2}$
$ \cong 1.604$
$\Rightarrow$ Moles of PCl5 = 3 $-$ 1.604 $\cong$ 1.396