Chemical Equilibrium
For a reaction at equilibrium
A(g) $\rightleftharpoons$ B(g) + ${1 \over 2}$ C(g)
the relation between dissociation constant (K), degree of dissociation ($\alpha$) and equilibrium pressure (p) is given by :
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
Explanation:
At 298 K : in aq. solution $[{H_3}{O^ + }][O{H^ - }] = {10^{ - 14}}$
$[{H_3}{O^ + }] = {{{{10}^{ - 14}}} \over {{{10}^{ - 2}}}} = {10^{ - 12}}$
A + B $\rightleftharpoons$ 2C
the value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is x $\times$ 10$-$1 M. The value of x is ____________. (Nearest integer)
Explanation:
$K = {{[C]_{eq}^2} \over {{{[A]}_{eq}}{{[B]}_{eq}}}} = {{{{(1 + 2x)}^2}} \over {(1 - x)(1 - x)}}$
$100 = {\left( {{{1 + 2x} \over {1 - x}}} \right)^2}$
$\left( {{{1 + 2x} \over {1 - x}}} \right) = 10$
$x = {3 \over 4}$
$[C]{e_{q.}} = 1 + 2x$
$ = 1 + 2\left( {{3 \over 4}} \right)$
= 2.5 M
= 25 $\times$ 10-1 M
N2O4(g) $\rightleftharpoons$ 2NO2(g) at 288 K is 47.9. The KC for this reaction at same temperature is ____________. (Nearest integer)
(R = 0.083 L bar K$-$1 mol$-$1)
Explanation:
For the equilibrium reaction
$ \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) $
the relationship between $ K_P $ and $ K_C $ is given by the equation:
$ K_P = K_C (RT)^{\Delta n} $
where $ \Delta n $ is the change in the number of moles of gas, $ R $ is the ideal gas constant, and $ T $ is the temperature in Kelvin.
For the given reaction:
$ \Delta n = 2 - 1 = 1 $
Given:
$ K_P = 47.9 $
$ R = 0.083 \, \text{L bar K}^{-1} \text{mol}^{-1} $
$ T = 288 \, \text{K} $
Substitute these values into the equation:
$ 47.9 = K_C (0.083 \times 288)^{1} $
$ K_C = \frac{47.9}{0.083 \times 288} $
Calculate:
$ 0.083 \times 288 = 23.904 $
$ K_C = \frac{47.9}{23.904} $
$ K_C \approx 2.0033 $
Rounding to the nearest integer, the value of $ K_C $ is
$ \boxed{2} $
In an equilibrium mixture, the partial pressures are
PSO3 = 43 kPa; PO2 = 530 Pa and PSO2 = 45 kPa. The equilibrium constant KP = ___________ $\times$ 10$-$2. (Nearest integer)
Explanation:
Given values are : pSO3 = 45kPa, pSO2 = 530 Pa = 0.53 kPa
pSO2 = 43 kPa
Now, ${K_p} = {{{{[{p_{S{O_3}(g)}}]}^2}} \over {{{[{p_{S{O_2}(g)}}]}^2} \times [{p_{{O_2}}}]}}$
On putting given values, we get
$ \Rightarrow {K_p} = {{{{(43)}^2}} \over {{{(45)}^2} \times 0.53}}$
$ = {{1849} \over {2025 \times 0.53}} = {{1849} \over {1073.25}}$
$ = 1.7228$
$ = {{1.7228 \times {{10}^2}} \over {{{10}^2}}} = 172.28 \times {10^{ - 2}} = 172$
Hence, the equilibrium constant, Kp = 172.
The equilibrium constant KC for this reaction is ________ $\times$ 10$-$2. (Round off to the Nearest Integer).
[Use : R = 8.3 J mol$-$1 K$-$1, ln 10 = 2.3 log10 2 = 0.30, 1 atm = 1 bar]
[antilog ($-$0.3) = 0.501]
Explanation:
$\Delta$G$^\circ$ = $-$ RTln(Kp)
$ \Rightarrow $ 25.2 $\times$ 103 = $-$8.3 $\times$ 400 $\times$ 2.3 log (Kp)
$ \Rightarrow $ Kp = 10$-$3.3
= 10$-$3 $\times$ 0.501
= 5.01 $\times$ 10$-$4 Bar$-$1
Also,
${{{K_p}} \over {{K_c}}} = {(RT)^{\Delta {n_g}}}$
$ \Rightarrow {{{K_p}} \over {{K_c}}} = {(RT)^{ - 1}}$
$ \Rightarrow {K_c} = {K_p}(RT)$
$ = 5.01 \times {10^{ - 4}} \times 8.3 \times 400$
$ = 1.66 \times {10^{ - 5}}$ m3/mole
$ = 1.66 \times {10^{ - 2}}$ L/mol
$N_{2}O_{4}\left( g\right) \rightleftharpoons 2NO_{2}\left( g\right) $
The temperature at which KC = 20.4 and KP = 600.1, is ____________ K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K$-$1 mol$-$1]
Explanation:
$\Delta$ng = 2 $-$ 1 = 1
KP = KC(RT)$\Delta$ng
600.1 = 20.4 (0.0831 $\times$ T)1
$ \Rightarrow $ T = ${{600.1} \over {20.4 \times 0.0831}}$ = 354 K
[Neglect volume change on adding HA. Assume degree of dissociation <<1 ]
Explanation:
Now,
${K_a} = {{[{H^ + }][{A^ - }]} \over {[HA]}}$
$ \Rightarrow 2 \times {10^{ - 6}} = {{(0.1)({{10}^{ - 2}}\alpha )} \over {{{10}^{ - 2}}}}$
$ \Rightarrow \alpha = 2 \times {10^{ - 5}}$
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.
(Round off to the Nearest Integer). [R = 8.314 J mol$-$1 K$-$1; ln 10 = 2.303]
Explanation:
$-$9.478 $\times$ 103 = $-$495 $\times$ 8.314 ln Keq
ln Keq = 2.303 = ln 10
So, Keq = 10
Now, A(g) $\rightleftharpoons$ B(g)
$\matrix{ {t = 0} & {22} & 0 \cr {t = t} & {22 - x} & x \cr } $
$Keq = {{[B]} \over {[A]}} = {x \over {(22 - x)}} = 10$
x = 20
So, millimoles of B = 20
[R = 0.08206 dm3atm K$-$1mol$-$1]
Explanation:
[p = Total pressure at equilibrium = 1.9 atm]
Now, at equilibrium pV = (1 + 2x)RT
$ \Rightarrow 1 + 2x = {{pV} \over {RT}} = {{1.9 \times 25} \over {0.082 \times 300}} = 1.93$
[V = 25 L, R = 0.082 L atm mol$-$1 K$-$1 T = 300 K]
$ \Rightarrow x = {{1.93 - 1} \over 2} = 0.465$
$ \Rightarrow {K_p} = {{{p_A} \times p_B^2} \over {{p_{A{B_2}}}}} \Rightarrow {{\left( {{x \over {1 + 2x}}p} \right) \times {{\left( {{{2x} \over {1 + 2x}}p} \right)}^2}} \over {\left( {{{1 - x} \over {1 + 2x}}p} \right)}}$
$ = {{4{x^3} \times {p^3}} \over {{{(1 + 2x)}^3}}} \times {{(1 + 2x)} \over {(1 - x) \times p}} = {{4{x^3} \times {p^2}} \over {{{(1 + 2x)}^2} \times (1 - x)}}$
$ = {{4 \times {{(0.465)}^3} \times {{(1.9)}^2}} \over {{{(1 + 2 \times 0.465)}^2} \times (1 - 0.465)}} = 0.7285$ atm
$ = 72.85 \times {10^{ - 2}}$ atm $ \simeq 73 \times {10^{ - 2}} = x \times {10^{ - 2}}$
$\therefore$ $x = 73$
The value of stability constants K1, K2, K3 and K4 are 104, 1.58 x 103, 5 x 102 and 102 respectively.
The overall equilibrium constants for dissociation of ${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$ is x $ \times $ 10-12.
The value of x is ________. (Rounded off to the nearest integer)
Explanation:
K1 = 104
K2 = 1.58 $\times$ 103
K3 = 5 $\times$ 102
K4 = 102
Cu2+ + NH3 $\buildrel {K_1} \over \rightleftharpoons $ [Cu(NH3)]2+ .... (i)
[Cu(NH3)]2+ + NH3 $\buildrel {K_2} \over \rightleftharpoons $ [Cu(NH3)2]2+.... (ii)
[Cu(NH3)2]2+ + NH3 $\buildrel {K_3} \over \rightleftharpoons $ [Cu(NH3)3]2+..... (iii)
[Cu(NH3)3]2+ + NH3 $\buildrel {K_4} \over \rightleftharpoons $ [Cu(NH3)4]2+ ..... (iv)
On adding Eqs. (i), (ii), (iii) and (iv), we get
Cu2+ + 4NH3 $\buildrel {K} \over \rightleftharpoons $ [Cu(NH3)4]2+
$\therefore$ The overall reaction constant (k) or equilibrium constant for formation of [Cu(NH3)4]2+ is
K = K1 $\times$ K2 $\times$ K3 $\times$ K4
$ \Rightarrow $ K = 104 $\times$ 1.58 $\times$ 103 $\times$ 5 $\times$ 102 $\times$ 102
$ \Rightarrow $ K = 7.9 $\times$ 1011
where, K = equilibrium constant for formation of [Cu(NH3)4]2+
So, equilibrium constant 'K' for dissociation of [Cu(NH3)4]2+ is ${1 \over K}$.
$K' = {1 \over K} = {1 \over {7.9 \times {{10}^{11}}}} = 1.26 \times {10^{ - 12}}$
Hence, K' = x $\times$ 10$-$12
x = 1.26
The value of x is _______. (Rounded off to the nearest integer)
Explanation:
Let moles of both of Cl2 and Cl molecule be x.
Partial pressure of Cl is, ${p_{Cl}} = {x \over {2x}} \times 1 = {1 \over 2}$
Partial pressure of Cl2 is, ${p_{C{l_2}}} = {x \over {2x}} \times 1 = {1 \over 2}$
Now, ${K_p} = {{{{({p_{Cl}})}^2}} \over {{p_{C{l_2}}}}} $
$\Rightarrow {K_p} = {{{{(1/2)}^2}} \over {1/2}} = {1 \over 2} = 0.5$
= 5 $\times$ 10$-$1
Hence, x $\times$ 10$-$1
x = 5
[R = 8.31 J mol–1K-1 and ln 10 = 2.3)
Explanation:
Given, Kp (equilibrium constant) = 100
Temperature = 300 K
Pressure = 1 atm
Formula used, $\Delta$G$^\circ$ = $-$ RT ln Kp .... (i)
Here, $\Delta$G$^\circ$ = standard Gibb's free energy
R = gas constant = 8.31 J mol$-$1 K$-$1
Put value in Eq. (i), we get
$\Delta$G$^\circ$ = $-$ R (300) ln 100
$\Delta$G$^\circ$ = $-$ R (300) (2) ln (10)
$\because$ ln (10) = 2.3
$\Delta$G$^\circ$ = $-$ R(300) (2) (2.3)
$\Delta$G$^\circ$ = $-$ 1380 R
Hence, $\Delta$G$^\circ$ = $-$ xR
$ \therefore $ x = 1380
N2(g) + 3H2(g) ⇌ 2NH3(g)
The value of KC for the following reaction is :
NH3(g) ⇌ ${1 \over 2}$N2(g) + ${3 \over 2}$H2(g)
Fe2N(s) + ${3 \over 2}$H2(g) ⇌ 2Fe(s) + NH3(g)
| Temperature | Equilibrium Constant |
|---|---|
| T1 = 25oC | K1 = 10 |
| T2 = 100oC | K2 = 100 |
The values of $\Delta $Ho, $\Delta $Go at
T1 and $\Delta $Go at T2 (in kJ mol–1) respectively, are close to :
[Use R = 8.314 J K–1 mol–1]
N2O4(g) ⇌ 2NO2(g); $\Delta $Ho = +58 kJ
For each of the following cases (a, b), the direction in which the equilibrium shifts is :
(a) Temperature is decreased.
(b) Pressure is increased by adding N2 at constant T.
A ⇌ B + C is $K_{eq}^{(1)}$ and that of
B + C ⇌ P is $K_{eq}^{(2)}$, the equilibrium
constant for A ⇌ P is :
of Y and 0.5 mol of Z were taken in a 1 L vessel and
allowed to react. At equilibrium, the concentration
of Z was 1.0 mol L–1. The equilibrium constant of reaction
is ${x \over {15}}$. The value of x is _________.
Explanation:
Keq = ${{{{\left( 1 \right)}^2}} \over {{3 \over 4} \times {5 \over 4}}}$ = ${{16} \over {15}}$
$ \therefore $ x = 16
2SO2(g) + O2(g) = 2SO3(g), $\Delta $H = –57.2 kJ mol–1 and KC = 1.7 × 1016
Which of the following statement is incorrect ?
S(s) + O2(g) ⇋ SO2(g); K1 = 1052
2S(s) + 3O2(g) ⇋ 2SO3(g); K2 = 10129
The equilibrium constant for the reaction,
2SO2(g) + O2(g) ⇋ 2SO3(g) is :
The total pressure when both the solids dissociated simultaneously is -
the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is -
N2(g) + 3H2(g) $\rightleftharpoons$ 2NH3(g)
The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that PNH3 << Ptotal at equilibrium)
N2(g) + O2(g) $\rightleftharpoons$ 2 NO(g)
N2O4(g) $\rightleftharpoons$ 2 NO(g)
N2(g) + 3H2(g) $\rightleftharpoons$ 2 NH3(g)
The relation between K1 and K2 is :
CO + Cl2 $\rightleftharpoons$ COCl2
At equilibrium, if one mole of CO is present then equilibrium constant (Kc) for the reaction is :