Chemical Equilibrium
[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
At $60^{\circ} \mathrm{C}$, dinitrogen tetroxide is dissociated. Find it's standard free energy change at this temperature and one atmosphere. [Given $\log 1.33=0.1239$]
Le-Chatelier's principle is not applicable to
Using the data provided, find the value of equilibrium constant for the following reaction at $298 \mathrm{~K}$ and $1 \mathrm{~atm}$ pressure.
$\begin{aligned} \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons & \mathrm{NO}_2(g) \\ \Delta_f H \mathrm{Y}[\mathrm{NO}(g)] & =90.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_f H \mathrm{Y}\left[\mathrm{NO}_2(g)\right] & =32.48 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta S Y a t ~298 \mathrm{~K} & =-70.8 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}$
$[\operatorname{antilog}(0.50)=3162 \text { ] }$
Standard entropies of $X_2, Y_2$ and $X Y_3$ are 60, 40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ respectively. At what temperature, the following reaction will be at equilibrium? [given: $\Delta H \Upsilon=-30 \mathrm{~kJ}$]
$\frac{1}{2} X_2+\frac{3}{2} Y_2 \rightleftharpoons X Y_3$
For the reaction $\mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{SO}_3(g)$, the percentage yield of product at different pressure is shown in the figure. Then, which among the following is true?
Which among the following denotes the correct relationship between $K_p$ and $K_c$ for the reaction, $2 A(g) \rightleftharpoons B(g)+C(g)$
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
A $\rightleftharpoons $ B
at 1000 K. At time t', the temperature of the system was increased to 2000 K and the system was allowed to reach equilibrium. Throughout this experiment the partial pressure of A was maintained at 1 bar. Given, below is the plot of the partial pressure of B with time. What is the ratio of the standard Gibbs energy of the reaction at 1000 K to that at 2000 K?
Explanation:
Using $\Delta G = \Delta {G^o} + RT\ln {K_p}$
At equilibrium : $\Delta {G^o} = - RT\ln {K_p}$
$\Delta G_1^0 = - R{T_1}\ln {K_{p1}}$ ... (i)
$\Delta G_2^0 = - R{T_2}\ln {K_{p2}}$ ...(ii)
From Eqs. (i) and (ii),
${{\Delta G_1^o} \over {\Delta G_2^o}} = {{{T_1}} \over {{T_2}}} \times {{\ln K{p_1}} \over {\ln K{p_2}}}$
$ = {{1000} \over {2000}} \times {{\ln (10)} \over {\ln (100)}} = {1 \over 4} = 0.25$
Aqueous solution of ferric nitrate when mixed with aqueous solution of potassium thiocyanate gives red colour solution. The intensity of red colour becomes constant on attaining equilibrium.
Choose the correct statement when the following chemical is added to the above solution at equilibrium.
I. Oxalic acid
II. Mercuric chloride
Both (I) and (II) will decrease the intensity of red colour.
Both (I) and (II) will increase the intensity of red colour.
(I) will increase but (II) will decrease the intensity of red colour.
(I) will decrease but (II) will increase the intensity of red colour.
For a given reaction, $2 A \rightleftharpoons B+C$, the equilibrium constant is $2 \times 10^{-3}$. If at any given time the composition of the reaction mixture is $[A]=[B]=[C]=6 \times 10^{-5} \mathrm{M}$; predict in which direction the reaction will proceed and the correct value for reaction quotient.
Forward direction and 1.0
Backward direction and 1.0
Forward direction and $3 \times 10^{-5}$
Backward direction and $3 \times 10^{-5}$
For a reversible reaction $A \rightleftharpoons B$, pre-exponential factor is same for both the forward and backward reactions and has value of $20 \mathrm{~S}^{-1}$. If the enthalpy change along the forward reaction is $-41.5 \mathrm{~kJ} / \mathrm{mol}$, the value of equilibrium constant at 500 K is
$e^{10}$
$e^9$
$e^8$
$e^7$
The vapour density of $\mathrm{N}_2 \mathrm{O}_4$ in $\mathrm{N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2$ is 40 . The degree of dissociation is
1.25
2.50
1.50
0.15
What is the equilibrium constant $\left(K_C\right)$ for the given reaction?
$ \mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} $
Where the equilibrium concentration of $\mathrm{N}_2, \mathrm{O}_2$ and NO are found to be $4 \times 10^{-3}, 3 \times 10^{-3}$ and $3 \times 10^{-3} \mathrm{M}$ respectively.
0.750
0.622
$9 \times 10^{-3}$
$12.8 \times 10^{-6}$
1.1 mole of A mixed with 2.2 moles of B and the mixture is kept in a 1 L flask and the equilibrium, A + 2B $\rightleftharpoons$ 2C + D is reached. If at equilibrium 0.2 mole of C is formed then the value of Kc will be
Sulphuric acid is a dibasic acid. It ionises in two stages and hence has two dissociation constants Ka1 and Ka2. Which of the following is the correct observation regarding Ka1 and Ka2 ?
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)
