Chemical Equilibrium
Pre-exponential factors for the forward and backward reactions are $10^{15} \mathrm{~s}^{-1}$ and $10^{11} \mathrm{~s}^{-1}$, respectively. If the value of $\log K$ for the reaction at $500 \mathrm{~K}$ is 6 , the value of $\left|\log k_b\right|$ at $250 \mathrm{~K}$ is ______.
$ \begin{aligned} & {[K=\text { equilibrium constant of the reaction }} \\\\ & k_f=\text { rate constant of forward reaction } \\\\ & \left.k_b=\text { rate constant of backward reaction }\right] \end{aligned} $
Explanation:
$\log \mathrm{k}_{\mathrm{f}}=\frac{-\mathrm{E}_{\mathrm{f}}}{2.303 \mathrm{RT}}+\log \mathrm{A}_{\mathrm{f}}$ [Arrhenius equation for forward reaction]
From plot when, $\frac{1}{\mathrm{~T}}=0.002, \log \mathrm{k}_{\mathrm{f}}=9$
$\begin{aligned} & \Rightarrow 9=\frac{-E_f}{2.303 R}(0.002)+\log \left(A_f\right) \\\\ & \text { Given }: A_f=10^{15} \mathrm{~s}^{-1} \\\\ & \Rightarrow 9=\frac{-E_f}{2.303 R}(0.002)+15 \\\\ & \Rightarrow \frac{E_f}{2.303 R}=\frac{6}{0.002}=3000\end{aligned}$
$\begin{aligned} \text { Now, } \mathrm{K} & =\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{b}}}=\frac{\mathrm{A}_{\mathrm{f}}}{\mathrm{A}_{\mathrm{b}}} \mathrm{e}^{-\left(\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}\right) / \mathrm{RT}} \\\\ \log \mathrm{K} & =-\frac{1}{2.303} \frac{\left(\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}\right)}{R T}+\log \left(\frac{10^{15}}{10^{11}}\right)\end{aligned}$
At $500 \mathrm{~K}$
$ \begin{aligned} & \Rightarrow 6=\frac{-\left(\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{b}}\right)}{500 \mathrm{R}(2.303)}+4 \\\\ & \Rightarrow(1000 \mathrm{R})(2.303)=\mathrm{E}_{\mathrm{b}}-\mathrm{E}_{\mathrm{f}} \\\\ & \Rightarrow(1000 \mathrm{R})(2.303)=\mathrm{E}_{\mathrm{b}}-3000(2.303 \mathrm{R}) \\\\ & \Rightarrow \mathrm{E}_{\mathrm{b}}=4000 \mathrm{R}(2.303) ..........(i) \end{aligned} $
Now $\mathrm{k}_{\mathrm{b}}=\mathrm{A}_{\mathrm{b}} \mathrm{e}^{-\mathrm{E}_{\mathrm{b}} / \mathrm{RT}}$
$\begin{aligned} & \Rightarrow \log \mathrm{k}_{\mathrm{b}}=\frac{-\mathrm{E}_{\mathrm{b}}}{2.303 \mathrm{RT}}+\log \mathrm{A}_{\mathrm{b}} \\\\ & \text { At } 250 \mathrm{~K}, \\\\ & \Rightarrow \log \mathrm{k}_{\mathrm{b}}=-\frac{4000}{250}+\log \left(10^{11}\right) \text { [From equation (1)] } \\\\ & \quad=-16+11=-5 \\\\ & \left|\log \mathrm{k}_{\mathrm{b}}\right|=5\end{aligned}$
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$
Fe2+(aq) + S2-(aq) ⇌ FeS(s)
When equal volumes of
0.06 M Fe2+(aq) and 0.2 M S2$ - $(aq)
solutions are mixed, the equilibrium concentration of Fe2+(aq) is found by Y $ \times $ 10$ - $17 M. The value of Y is .................
Explanation:
[Here, Kc >>103, thus limiting reagent will be consumed almost completely, 0.03 $ - $ X = 0 $ \therefore $ X = 0.03]
From equilibrium constant,
${K_c} = {{[FeS]} \over {[F{e^{2 + }}][{S^{2 - }}]}}$
${K_c} = {1 \over {X \times 0.07}}$
[For $\mathop {Fes(s)}\limits_{(Pure\,solid)} = 1\,mol\,{L^{ - 1}}]$]
$1.6 \times {10^{17}} = {1 \over {X \times 0.07}}$
$X = {1 \over {1.6 \times {{10}^{17}} \times 0.07}}$
$ = 8.9 \times {10^{ - 17}}$
Given, X = Y $ \times $ 10-17 = 8.9 $ \times $ 10-17
$ \therefore $ Y = 8.9
Explanation:
Given p1 = 5 bar, V1 = 1 m3, T1 = 400 K
So, ${n_1} = {5 \over {400R}}$ (from pV = nRT)
Similarly, p2 = 1 bar, V2 = 3 m3, T2 = 300 K,
${n_2} = {3 \over {300R}}$
Let at equilibrium the new volume of A will be (1 + x)
So, the new volume of B will be (3 $-$ x)
Now, from the ideal gas equation,
${{{p_1}{V_1}} \over {{n_1}R{T_1}}} = {{{p_2}{V_2}} \over {{n_2}R{T_2}}}$
and at equilibrium (due to conduction of heat)
${{{p_1}} \over {{T_1}}} = {{{p_2}} \over {{T_2}}}$
So, ${{{V_1}} \over {{n_1}}} = {{{V_2}} \over {{n_2}}}$ or ${V_1}{n_2} = {V_2}{n_1}$
After putting the values
$(1 + x) \times {3 \over {300R}} = (3 - x) \times {5 \over {400R}}$
or $(1 + x) = {{(3 - x)5} \over 4}$ or $4(1 + x) = 15 - 5x$
or $4 + 4x = 15 - 5x$ or $x = {{11} \over 9}$
Hence, new volume of A i.e., (1 + x) will comes as $1 + {{11} \over 9} = {{20} \over 9}$ or 2.22.
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equations:
X2 (g) $\leftrightharpoons$ 2X (g)
The standard reaction Gibbs energy, $\Delta _rG^o$, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by $\beta$. Thus, $\beta _{equilibrium}$ is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given R = 0.083 L bar K-1 mol-1)
Question
The INCORRECT statement among the following for this reaction, is
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equations:
X2 (g) $\leftrightharpoons$ 2X (g)
The standard reaction Gibbs energy, $\Delta _rG^o$, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by $\beta$. Thus, $\beta _{equilibrium}$ is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given R = 0.083 L bar K-1 mol-1)
Question
The equilibrium constant Kp for this reaction at 298 K, in terms of $\beta _{equilibrium}$, is
CaCO3(s) $\leftrightharpoons$ CaO(s) + CO2(g).
For this equilibrium, the correct statement(s) is (are)
The equilibrium
$2C{u^+} \to Cu^\circ + C{u^{2+}}$
In aqueous medium at 25$^\circ$C shifts towards the left in the presence of
Statement 1 : For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero.
and
Statement 2 : At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy.
$ \begin{aligned} & \mathrm{Ag}^{+}+\mathrm{NH}_3 \quad\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)\right]^{+} \\ & k_1=3.5 \times 10^{-3} \\ & {\left[\mathrm{Ag}\left(\mathrm{NH}_3\right]^{+}+\mathrm{NH}_3 \quad\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}\right.} \end{aligned} $
$k_2=1.7 \times 10^{-3}$, then the formation constant of $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$ is :
$6.08 \times 10^{-6}$
$6.08 \times 10^6$
$6.08 \times 10^{-9}$
None of these
$ \mathrm{N}_2+3 \mathrm{H}_2 \to 2 \mathrm{NH}_3 $
Which is the correct statement if $\mathrm{N}_2$ is added at equilibrium condition?
The equilibrium will shift to forward direction because according to second law of thermodynamics, the entropy must increase in the direction of spontaneous reaction.
The condition for equilibrium is $\mathrm{G}_{\mathrm{N}_2}+3 \mathrm{G}_{\mathrm{H}_2} \quad 2 \mathrm{G}_{\mathrm{NH}_3}$ where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent.
The catalyst will increase the rate of forward reaction by and that of backward reaction by $\beta$.
Catalyst will not alter the rate of either of the reaction.
Explanation:
The reaction involved is NH4HS(g) $\rightleftharpoons$ NH3(g) + H2S(g).
Mass of ammonium hydrogen sulphide decomposed $ = {{3.06 \times 30} \over {100}} = 0.918$ g
Moles of NH4HS decomposed $ = {{0.918} \over {51}} = 0.018$
Moles of ammonia formed = 0.018 ; Moles of hydrogen sulphide formed = 0.018
Thus, $[N{H_3}] = {{0.018} \over 2} = 0.009$ M ; $[{H_2}S] = {{0.018} \over 2} = 0.009$ M
Applying the law of chemical equilibrium on the reaction involved.
${K_c} = {{[N{H_3}(g)][{H_2}s(g)]} \over {[N{H_4}HS]}} = {{0.009 \times 0.009} \over 1}$ (Molar conc. of solid is equal to one)
$ = 8.1 \times {10^{ - 5}}$
$\therefore$ ${K_p} = {K_c} \times {(RT)^{\Delta n}}$
Here R = 0.081 lit atm K$-$1 mol$-$1 , T = 27$^\circ$C = 300 K ; $\Delta$n = 2
Substituting the values, we get ${K_c} = 8.1 \times {10^{ - 5}} \times {(0.081 \times 300)^2} = 4.78 \times {10^{ - 2}}$