$\Delta$vap H $-$ $\Delta$vap U = _____________ $\times$ 102 J mol$-$1. (Round off to the Nearest Integer)
[Use : R = 8.31 J mol$-$1 K$-$1]
[Assume volume of H2O(l) is much smaller than volume of H2O(g). Assume H2O(g) treated as an ideal gas]
Explanation:
$\Delta$H = $\Delta$U + $\Delta$ngRT
For 1 mole waters;
$\Delta$ng = 1
$\therefore$ $\Delta$ngRT = 1 mol $\times$ 8.31 J/mol-k $\times$ 373 K
= 3099.63 J $ \cong $ 31 $\times$ 102 J
Explanation:
$\Delta$U = 150 $-$ 200 = $-$50 J
Magnitude = 50 J = |$\Delta$U |
Explanation:
The enthalpy of sublimation is the total amount of energy required to convert a solid directly into a gas. This can be calculated by summing the enthalpy of fusion (solid to liquid) and the enthalpy of vaporization (liquid to gas). Mathematically, this relationship is represented as:
$ \Delta H_{\text{sublimation}} = \Delta H_{\text{fusion}} + \Delta H_{\text{vaporization}} $
Given:
Enthalpy of fusion, $\Delta H_{\text{fusion}} = 2.8 \, \text{kJ mol}^{-1}$
Enthalpy of vaporization, $\Delta H_{\text{vaporization}} = 98.2 \, \text{kJ mol}^{-1}$
Substitute these values into the equation:
$ \Delta H_{\text{sublimation}} = 2.8 \, \text{kJ mol}^{-1} + 98.2 \, \text{kJ mol}^{-1} = 101.0 \, \text{kJ mol}^{-1} $
Therefore, the enthalpy of sublimation of the substance (X) is approximately 101 kJ mol$-1$.
Explanation:
For 1 g of graphite = ${{248} \over {12}}$ = 20.67 kJ/gm heat evolved.
Explanation:
$\Delta$H = 51.4 kJ/mol
$\Delta$G = $\Delta$H $-$ T$\Delta$S
$-$49400 = 51400 $-$ 300$\Delta$S
$\Delta S = {{ + 100800} \over {300}} = 336$ JK$-$1 mol$-$1
the reaction enthalpy $\Delta$rH = __________ kJ mol$-$1. (Round off to the Nearest Integer).
[ Given : Bond enthalpies in kJ mol$-$1 : C-C : 347, C = C : 611; C-H : 414, H-H : 436 ]
Explanation:
$\Delta H = {E_{C - C}} + 6{E_{C - H}} - {E_{C = C}} - 4{E_{C - H}} - {E_{H - H}}$
$ = 347 + 6(414) - 611 - 4 \times 414 - 436$
$ = 347 + 828 - 1047$
$ = 128$ KJ/ mol
For the reaction
3CaO + 2Al $ \to $ 3Ca + Al2O3 the standard reaction enthalpy $\Delta$rH0 = _________ kJ.
(Round off to the Nearest Integer)
Explanation:
$\Delta H_{reaction}^o = 3\Delta {H^o}_f(Ca,s) + \Delta {H^o}_f(A{l_2}{O_3},s) - 3\Delta {H^o}_f(CaO,s) - 2\Delta {H^o}_f(Al,S)$
$ = 0 + ( - 1675) - 3( - 635) - 0$
$ = - 1675 + 1905$
$ = 230$ KJ
[Given : R = 8.314 J mol$-$1 K$-$1. Assume, hydrogen is an ideal gas] [Atomic mass of Fe is 55.85 u]
Explanation:
${{50} \over {55.85}}moles$
Moles of Fe = ${{50} \over {55.85}}moles$ = Moles of H2
No. of H2 produced $ = {{50} \over {55.85}}moles$
Work done $ = - {P_{ext}}\,.\,\Delta V$
$ = - \Delta {n_g}RT$
$ = - {{50} \over {55.85}} \times 8.314 \times 298$
= -2218.05 J
Nearest integer = 2218
[Given : The values of standard enthalpy of formation of SF6(g), S(g) and F(g) are - 1100, 275 and 80 kJ mol$-$1 respectively.]
Explanation:
So, $\Delta_f H^{\circ}[S, g]+6 \times \Delta_f H^{\circ}[F, g]=\Delta_f H^{\circ}\left[SF_6, g\right]$ $+ 6 \times E_{S-F}$
$\left[\therefore E_{S-F}=\right.$ Average $S-\mathrm{F}$ bond energy in $\left.\mathrm{SF}_6\right]$
$275+6 \times 80=-1100+6 \times E_{S-F}$
$\Rightarrow E_{S-F}=\frac{275+6 \times 80+1100}{6}$ $=309.16 \mathrm{~kJ} \mathrm{~mol}^{-1}=309 \mathrm{~kJ} \mathrm{~mol} \mathrm{~m}^{-1}$
(${\Delta _r}{H^\Theta }$ = 80 kJ mol$-$1) the entropy change ${\Delta _r}{S^\Theta }$ depends on the temperature T (in K) as ${\Delta _r}{S^\Theta }$ = 2T (J K$-$1mol$-$1).
Minimum temperature at which it will become spontaneous is ___________ K. (Integer)
Explanation:
For a reaction to be spontaneous
$\Delta G^\circ < 0$
$ \Rightarrow $ $\Delta H^\circ - T\Delta S^\circ < 0$
$ \Rightarrow $ $T > {{\Delta H^\circ } \over {\Delta S^\circ }}$
$ \Rightarrow $ $T > {{80000} \over {2T}}$
$ \Rightarrow $ 2T2 > 80000
$ \Rightarrow $ T2 > 40000
$ \Rightarrow $ T > 200
The minimum temperature to make it spontaneous is 200 K.
Explanation:
$\Delta $E = (Ea)f – (Ea)b
$ \Rightarrow $ – 20 = 30 – (Ea)b
$ \Rightarrow $ (Ea)b = 50 kJ
Explanation:
The gas performs isothermal irreversible work (W).
where, $\Delta$U = 0 (change in internal energy)
From, 1st law of thermodynamics,
$\Rightarrow$ $\Delta$U = $\Delta$Q + W
$\Rightarrow$ 0 = $\Delta$Q + W
$\Rightarrow$ $\Delta$Q = $-$W
Now, $W = - {p_{ext}}({V_2} - {V_1})$
$ = - {p_{ext}}\left( {{{nRT} \over {{p_2}}} - {{nRT} \over {{p_1}}}} \right) = - {p_{ext}} \times nRT\left( {{1 \over {{p_2}}} - {1 \over {{p_1}}}} \right)$
Given, pext = 4.3 MPa, p1 = 2.1 MPa, p2 = 1.3 MPa, n = 5 mol, T = 293 K and R = 8.314 J mol$-$1 K$-$1
$ = - 4.3 \times 5 \times 8.314 \times 293\left( {{1 \over {1.3}} - {1 \over {2.1}}} \right)$
= $-$ 15347.70 J mol$-$1
= $-$ 15.347 kJ mol$-$1 $ \simeq $ $-$ 15 kJ mol$-$1
$\Rightarrow$ $\Delta$Q = 15 kJ mol$-$1
$N{H_2}C{H_{(S)}} + {3 \over 2}{O_{2(g)}} \to {N_{2(g)}} + {O_{2(g)}} + {H_2}{O_{(I)}}$
is _________ kJ. (Rounded off to the nearest integer) [Assume ideal gases and R = 8.314 J mol$-$1 K$-$1]
Explanation:
$\Delta ng = (1 + 1) - {3 \over 2} = {1 \over 2}$
$\Delta H = \Delta U + \Delta ng\,RT$
$ = - 742.24 + {1 \over 2} \times {{8.314 \times 298} \over {1000}}$
$ = - 742.24 + 1.24$
$ = -741$ kJ/mol
Explanation:
$\Delta H = 495.8$
${1 \over 2}B{r_2}(l) + {e^ - }\buildrel {} \over \longrightarrow B{r^ - }(g)$
$\Delta H = 325$
$N{a^ + }(g) + B{r^ - }(g)\buildrel {} \over \longrightarrow NaBr(s)$
$\Delta H = - 728.4$
$Na(s) + {1 \over 2}B{r_2}(l)\buildrel {} \over \longrightarrow NaBr(s).$
$\Delta H = ?$
$\Delta H = 495.8 - 325 - 728.4 - 557.6$ kJ
$ = - 5576 \times {10^{ - 1}}$ kJ
$3HC \equiv C{H_{(g)}} \rightleftharpoons {C_6}{H_{6(l)}}$
[Given : ${\Delta _f}{G^o}(HC \equiv CH) = - 2.04 \times {10^5}$ J mol$-$1 ; ${\Delta _f}{G^o}({C_6}{H_6}) = - 1.24 \times {10^5}$ J mol$-$1 ; R = 8.314 J K-1 mol$-$1]
Explanation:
$\mathop {3HC \equiv CH(g)}\limits_{Acetylene} \to \mathop {{C_6}{H_6}(l)}\limits_{Benzene} $
Given, $\Delta G_f^o (CH \equiv CH) = - 2.04 \times {10^5}$ J mol$-$1
$\Delta G_f^o({C_6}{H_6}) = - 1.24 \times {10^5}$ J mol$-$1
Gibb's free energy, $\Delta G_f^o = - nRT\ln K$
$\Delta G_f^o = \sum {{{(\Delta G_F^o)}_P} - \sum {{{(\Delta G_F^o)}_R}} } $
$ - nRT\ln K = - n'RT\ln {K_p} - ( - n''RT\ln {K_f})$
$ \Rightarrow - RT\ln K = 1 \times ( - 1.24 \times {10^5}) - ( - 3 \times 2.04 \times {10^5}) $
$ \Rightarrow $ $- 2.303 \times R \times T\log K = 4.88 \times {10^5}$
$ \Rightarrow $ $\log K = - {{4.88 \times {{10}^5}} \over {2.303 \times 8.314 \times 273}}$
$ \Rightarrow $ $n\ln K = n'\ln {K_p} - ( - n''ln{K_f})$
$ \Rightarrow $ K = 85.52
$\Rightarrow$ K = 855 $\times$ 10$-$1
x = 855
2A(g) $ \to $ A2(g)
at 298 K, $\Delta $Uo = –20 kJ mol–1, $\Delta $So = –30 JK–1 mol–1,
then the $\Delta $Go will be _____ J.
Explanation:
= ($\Delta $Uo + $\Delta $ngRT) - T$\Delta $So
= $\left[ {\left\{ { - 20 + \left( 1 \right){{8.314} \over {1000}} \times 298} \right\} - {{298} \over {1000}} \times \left( { - 30} \right)} \right]$ kJ
= – 13.537572 kJ
= – 13537.57 Joule
Explanation:
$\Delta $H = –327 kcal; $\Delta $ng = – 1
$\Delta $H = $\Delta $U + $\Delta $ngRT
$ \Rightarrow $ $\Delta $U = – 327 + 2 × 10–3 × 300
= – 326.4 kcal
= – 326400 cal
$ \therefore $ Heat evolved = 326400 cal.
(Given : $\Delta $Hvap for water at 373 K = 41 kJ/mol,
R = 8.314 JK–1 mol–1)
Explanation:
90 gm of H2O = ${{90} \over {18}}$ moles of H2O = 5 moles of H2O
$\Delta $Hvap = $\Delta $U + $\Delta $ngRT
$ \Rightarrow $ $\Delta $U = $\Delta $Hvap - $\Delta $ngRT
= 41000 - 1$ \times $8.314$ \times $373
= 37898.878
For 5 moles, $\Delta $U = 37898.878 $ \times $ 5 = 189494.39 Joule
Explanation:
$ \Rightarrow $ 5000 = 4 × Cv(500 – 300)
$ \Rightarrow $ Cv = 6.25 JK–1mol–1
Explanation:
= $1\over2$ × (sum of parallel sides) × height
= $1\over2$ × (10+6) × 6
= $1\over2$ × 16 × 6
= 48 Joule
Note : Here pressure difference = 8 - 2 = 6 Pa and volume difference = 12 - 2 = 10
Explanation:
${\Delta _f}H$(C2H6) = 2$\Delta H$comb(Cgraphite) + 3$\Delta H$comb(H2)
- $\Delta H$comb(C2H6)
= – (286$ \times $2) - (393.5$ \times $3) -(-1560)
= -572 –1180.5+1560 = -192.5 kJ/mole
A($l$) $ \to $ 2B(g)
$\Delta U = 2.1\,kcal,\,\Delta S = 20\,cal\,{K^{ - 1}}$ at 300 K
Hence $\Delta $G in kcal is :
Explanation:
We know, $\Delta $H = $\Delta $U + $\Delta $ngRT
and $\Delta $G = $\Delta $H - T$\Delta $S
$ \therefore $ $\Delta $G = $\Delta $U + $\Delta $ngRT - T$\Delta $S
= 2.1 + ${{2 \times 2 \times 300} \over {1000}}$ - ${{300 \times 20} \over {1000}}$
= 2.1 + 1.2 - 6 = – 2.70 Kcal/mol
Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed completely under constant pressure at 300 K is ______.
Use: Universal gas constant (R) = 8.3 J K−1 mol−1; Atomic mass (in amu): H = 1, O = 16
Explanation:
Consider the following volume-temperature $(\mathrm{V}-\mathrm{T})$ diagram for the expansion of 5 moles of an ideal monoatomic gas.
Considering only $\mathrm{P}-\mathrm{V}$ work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence $\mathbf{X} \rightarrow \mathbf{Y} \rightarrow \mathbf{Z}$ is ____________.
[Use the given data: Molar heat capacity of the gas for the given temperature range, $\mathrm{C}_{\mathrm{V}, \mathrm{m}}=12 \mathrm{~J} \mathrm{~K}^{-1}$ $\mathrm{mol}^{-1}$ and gas constant, $\left.\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
Explanation:
For ideal gas
$ \begin{aligned} & \Delta H=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T} \\\\ & \because \mathrm{C}_{\mathrm{P}}=\mathrm{C}_{\mathrm{V}}+\mathrm{R}=12+8.3=20.3 \mathrm{~J} / \mathrm{K} \text {-mole } \\\\ & \therefore \Delta \mathrm{H}=5 \times 20.3 \times(415-335) \\\\ & \Delta \mathrm{H}=8120 \text { Joule } \end{aligned} $
[Use : $\ln 2=0.69$
Given : $S_\beta-S_\alpha=0$ at $0 \mathrm{~K}$ ]
Explanation:
$ \begin{aligned} & \mathrm{S}_{\alpha(600)}^{\mathrm{o}}=\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+\mathrm{C}_{\mathrm{P}(\alpha)} \ln \frac{600}{300} \\\\ & \mathrm{~S}_{\beta(600)}^{\mathrm{o}}=\mathrm{S}_{\beta(300)}^{\mathrm{o}}+\mathrm{C}_{\mathrm{P}(\beta)} \ell \mathrm{n} \frac{600}{300} \\\\ & \mathrm{~S}_{\beta(600)}^{\mathrm{o}}-\mathrm{S}_{\alpha(600)}^{\mathrm{o}}=\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+\left(\mathrm{C}_{\mathrm{P}(\beta)}-\mathrm{C}_{\mathrm{P}(\alpha)}\right) \ell \mathrm{n} 2 \\\\ & 6-5=\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+1 \times \ell \mathrm{n} 2 \\\\ & 1=\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+0.69 \end{aligned} $
So $\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}=0.31$
Explanation:
So at $600 \mathrm{~K} \quad \Delta \mathrm{G}_{\mathrm{rxn}}^{\circ}=0$ So $\Delta \mathrm{H}^{\circ}{ }_{\text {reaction (600) }}=\mathrm{T} \Delta \mathrm{S}^{\circ}{ }_{\text {reaction (600) }}$
$\Delta \mathrm{H}^{\circ}{ }_{(600)}=600 \times 1=600 \mathrm{Joule} / \mathrm{mole}$
So $\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=\Delta \mathrm{C}_{\mathrm{P}}\left(\mathrm{T}_2-\mathrm{T}_1\right)$
$\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=1 \times 300$
$\Delta \mathrm{H}_{300}=\Delta \mathrm{H}_{600}-300=600-300=300$ Joule $/$ mole.
$\mathrm{A} \rightarrow \mathrm{B}$ is an adiabatic process. If the total heat absorbed in the entire process $(\mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{B} \rightarrow \mathrm{C})$ is $\mathrm{R} T_2 \ln 10$, the value of $2 \log V_3$ is _______.
[Use, molar heat capacity of the gas at constant pressure, $C_{\mathrm{p}, \mathrm{m}}=\frac{5}{2} \mathrm{R}$ ]
Explanation:
$ \begin{aligned} &\Rightarrow \left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right) =\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\mathrm{V}-1} \\\\ &\Rightarrow\mathrm{~T}_1 \mathrm{~V}_1^{\gamma-1} =\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\\\ &\Rightarrow600(10)^{2 / 3} =60\left(\mathrm{~V}_2\right)^{2 / 3} \\\\ &\Rightarrow\left(\mathrm{~V}_2\right)^{2 / 3} =(10)^{5 / 3} \\\\ &\Rightarrow\mathrm{~V}_2 =(10)^{5 / 2} \\\\ &\Rightarrow\mathrm{Q}_{\mathrm{AB}} =0 \end{aligned} $
$\begin{aligned} \mathrm{Q}_{\mathrm{AC}} & =\mathrm{nRT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) \\\\ & =\mathrm{RT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) .........(i)\end{aligned}$
$\begin{gathered}\text { Total heat absorbed }=\mathrm{RT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) \\\\ =\mathrm{RT}_2 \ln (10) ........(ii)\end{gathered}$
Equating equation (i) and equation (ii)
$ \begin{aligned} &\mathrm{RT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) =\mathrm{RT}_2 \ln (10) \\\\ &\Rightarrow\ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) =\ln (10) \end{aligned} $
$ \mathrm{V}_3=10 \mathrm{~V}_2 $
Substitute value of $\mathrm{V}_2$
$ \begin{aligned} & =10(10)^{5 / 2}=(10)^{7 / 2} \\\\ V_3 & =(10)^{7 / 2} \end{aligned} $
Taking $\log$ on both side, we get
$ \begin{aligned} &\Rightarrow \log \left(\mathrm{V}_3\right) =\log (10)^{7 / 2} \\\\ &\Rightarrow \log \left(\mathrm{V}_3\right) =\frac{7}{2} \log (10) \\\\ &\Rightarrow 2 \log \left(\mathrm{V}_3\right) =7 \log (10) \\\\ \therefore 2 \log \left(\mathrm{V}_3\right) =7 \end{aligned} $
If $\mathrm{T}_1=2 \mathrm{~T}_2$ and $\left(\Delta \mathrm{G}_2^{\Theta}-\Delta \mathrm{G}_1^{\Theta}\right)=\mathrm{RT}_2 \ln \mathrm{x}$, then the value of $\mathrm{x}$ is _______.
$\left[\Delta \mathrm{G}_1^{\Theta}\right.$ and $\Delta \mathrm{G}_2^{\Theta}$ are standard Gibb's free energy change for the reaction at temperatures $\mathrm{T}_1$ and $\mathrm{T}_2$, respectively.]
Explanation:
$\Rightarrow$ At $\mathrm{T}_1 \mathrm{~K}: \mathrm{K}_{\mathrm{P}_1}=\frac{4}{2}=2$
$ \begin{array}{ll} \text { At } \mathrm{T}_2 \mathrm{~K}: & \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g}) \\\\ \mathrm{t}=0 & 6 \\\\ \mathrm{t}=\infty & 6-\mathrm{y} \quad \mathrm{y}=2 \text { (from plot) } \end{array} $
$ \Rightarrow \text { At } \mathrm{T}_2 \mathrm{~K}: \mathrm{K}_{\mathrm{P}_2}=\frac{2}{4}=\frac{1}{2} $
Since
$ \begin{aligned} \Delta \mathrm{G}_1^0 & =-\mathrm{RT}_1 \ln \mathrm{KT}_1 ............(i) \\\\ \Delta \mathrm{G}_2^0 & =-\mathrm{RT}_2 \ln \mathrm{KT}_2 ............(ii) \end{aligned} $
$\begin{aligned} \Delta \mathrm{G}_2^0-\Delta \mathrm{G}_1^0 & =-\mathrm{RT}_2 \ln \mathrm{KT}_2+\mathrm{RT}_1 \ln \mathrm{KT}_1 \\\\ & =-\mathrm{RT}_2 \ln \frac{1}{2}+\mathrm{RT}_1 \ln 2 \\\\ & =-\mathrm{RT}_2 \ln \frac{1}{2}+\mathrm{R}\left(2 \mathrm{~T}_2\right) \ln 2 \\\\ & =\mathrm{RT}_2 \ln 2+2 \mathrm{RT}_2 \ln 2 \\\\ & =\mathrm{RT}_2 \ln 2+2 \mathrm{RT}_2 \ln 2\end{aligned}$
$\begin{aligned} \Delta \mathrm{G}_2^0-\Delta \mathrm{G}_1^0 & =3 \mathrm{RT}_2 \ln 2 \\\\ & =\mathrm{RT}_2\left(\ln 2^3\right) \\\\ & =\mathrm{RT}_2 \ln 8\end{aligned}$
$2 \mathrm{~mol} \,\mathrm{of}\, \mathrm{Hg}(\mathrm{g})$ is combusted in a fixed volume bomb calorimeter with excess of $\mathrm{O}_{2}$ at $298 \mathrm{~K}$ and 1 atm into $\mathrm{HgO}(s)$. During the reaction, temperature increases from $298.0 \mathrm{~K}$ to $312.8 \mathrm{~K}$. If heat capacity of the bomb calorimeter and enthalpy of formation of $\mathrm{Hg}(g)$ are $20.00 \mathrm{~kJ} \mathrm{~K}^{-1}$ and $61.32 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$ at $298 \mathrm{~K}$, respectively, the calculated standard molar enthalpy of formation of $\mathrm{HgO}(s)$ at 298 $\mathrm{K}$ is $\mathrm{X}\, \mathrm{kJ}\, \mathrm{mol}^{-1}$. The value of $|\mathrm{X}|$ is _________ .
[Given: Gas constant $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]
Explanation:
$2Hg(g)+O_2(g) \longrightarrow 2HgO(s)$
Heat capacity of calorimeter = 20 kJ K$-1$
Rise in temperature = 14.8 K
$\Delta H^\circ=\Delta U^\circ+\Delta n_g RT$
$=-296-3\times8.3\times298\times10^{-3}$
$\simeq -303.42$ kJ
$\Delta H^\circ = \Delta H{^\circ _f}(HgO(s)) - \Delta H{^\circ _f}(Hg(g))$
$ - 303 - 42 = \Delta H{^\circ _f}(HgO(s)) - 2 \times 61.32$
$\Delta H{^\circ _f}(HgO(s)) = - 303.42 - 122.64$
$ = - 180.78$ kJ
$\left| {\Delta H{^\circ _f}(HgO(s))} \right| = 90.39$ kJ mol$-$1
(U : internal energy, S : entropy, p : pressure, V : volume, R : gas constant)
(Given : molar heat capacity at constant volume, CV,m of the gas is ${5 \over 2}$R)
Explanation:
$\therefore$ ${W_I} = \Delta U = n{C_{V,m}}\Delta T$
$ = - (2250 - 450)R = - 1800R$ .... (i)
$\Delta U = n{C_{V,m}}\Delta T$
$ - 1800R = 1 \times {{5R} \over 2} \times \Delta T$
$ \Rightarrow \Delta T = - 720K$
${T_2} - {T_1} = - 720K$
${T_2} = - 720K + 900K = 180K$
For process II,
${W_{II}} = - nR{T_2}\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$
$ = - 1 \times R \times 180\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$ .... (ii)
As, both work done for process I and II are equal,
Therefore, WI = WII
$ - 1800R = - R \times 180 \times \ln \left( {{{{V_3}} \over {{V_2}}}} \right)$
$\ln \left( {{{{V_3}} \over {{V_2}}}} \right) = {{1800} \over {180}} \Rightarrow \ln \left( {{{{V_3}} \over {{V_2}}}} \right) = 10$
Explanation:
$ \mathrm{X}(\mathrm{s}) \rightleftharpoons \mathrm{X}(\mathrm{s})+\mathrm{Z}(\mathrm{g}) $
We have
$ \frac{d(\ln K)}{d(1 / T)}=\frac{-\Delta H^{\circ}}{R} $
And $ K=\frac{p_{z}}{p^{\circ}} $
Substituting the value of $K$ in Eq. (1), we get
$ \frac{d\left(\ln \frac{p_{z}}{p^{\circ}}\right)}{d\left(\frac{1}{T}\right)}=\frac{-\Delta H^{\circ}}{R} $
Or
$ d\left(\ln \frac{p_{z}}{p^{\circ}}\right)=\frac{-\Delta H^{\circ}}{R} d\left(\frac{1}{T}\right) $
Integrating both sides of Eq. (2), we get
$ \ln \frac{p_{z}}{p^{\circ}}=\frac{-\Delta H^{\circ}}{R T}+\ln A $
Where $\ln A=$ Integration constant
From the given plot of $\ln \frac{p_{z}}{p^{\circ}} \mathrm{vs} \frac{10^{4}}{T}$
The slope of the plot is
$ \text { Slope }=\frac{-\Delta H^{\circ}}{R}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-7-(-3)}{\left(\frac{12}{10^{4}}-\frac{10}{10^{4}}\right)} $
$\frac{-\Delta H^{\circ}}{R}=\frac{-4}{2} \times 10^{4}$
$\Delta H^{\circ}=-2 \times 10^{4} \times R$
$\Delta H^{\circ}=2 \times 8.314 \times 10^{4}$
$\Delta H^{\circ}=16.628 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}$
$\Delta H^{\circ}=166.28 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
$ \mathrm{X}(\mathrm{s}) \rightleftharpoons \mathrm{X}(\mathrm{s})+\mathrm{Z}(\mathrm{g}) $
We have
$ \frac{d(\ln K)}{d(1 / T)}=\frac{-\Delta H^{\circ}}{R} $
And $ K=\frac{p_{z}}{p^{\circ}} $
Substituting the value of $K$ in Eq. (1), we get
$ \frac{d\left(\ln \frac{p_{z}}{p^{\circ}}\right)}{d\left(\frac{1}{T}\right)}=\frac{-\Delta H^{\circ}}{R} $
Or
$ d\left(\ln \frac{p_{z}}{p^{\circ}}\right)=\frac{-\Delta H^{\circ}}{R} d\left(\frac{1}{T}\right) $
Integrating both sides of Eq. (2), we get
$ \ln \frac{p_{z}}{p^{\circ}}=\frac{-\Delta H^{\circ}}{R T}+\ln A $
Where $\ln A=$ Integration constant
We can also calculate the intercept $(\ln A)$ from the plot of $\ln \frac{p_{z}}{p^{\circ}} \mathrm{vs} \frac{1}{T}$.
Now, from the equation of straight line, we have
or
$ \begin{aligned} &y=m x+C \\ &C=y-m x \end{aligned} $
From the plot, we have
$ m=-2 \times 10^{4} \mathrm{~K}, y=-3, x=\frac{10}{10^{4}} \mathrm{~K} $
Now, substituting the above values in Eq. (4), we get
$ \begin{aligned} &C=-3-\left(-2 \times 10^{4} \times \frac{10}{10^{4}}\right) \\ &C=-3+20 \\ &C=17 \end{aligned} $
$\Rightarrow C=17$
This intercept is equal to $\ln A$ in Eq. (3), so we have
$ \ln A=C=17 $
From the thermodynamic equations, we have
$ \Delta G^{\circ}=\Delta H^{\circ}-7 \Delta S^{\circ} $
For any equilibrium reaction, the Gibb's energy is,
$ \Delta G^{\circ}=-R T \ln K $
On comparing both the equations, we get
$ -R T \ln K=\Delta H^{\circ}-T \Delta S^{\circ} $
Or
$ \begin{aligned} \ln K &=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R} \\ \left(\frac{p_{z}}{p^{\circ}}\right) &=\frac{-\Delta H^{\circ}}{R T}+\frac{\Delta S^{\circ}}{R} \end{aligned} $
On comparing Eq. (3) and (6), we have
$ \frac{\Delta S^{\circ}}{R}=\ln A=17 $
Or
$ \begin{aligned} \Delta S^{\circ} &=17 R \\ &=17 \times 8.314 \\ &=141.34 \,\mathrm{JK}^{-1} \end{aligned} $
At $298K:{\Delta _f}H^\circ [Sn{O_2}(s)] = - 581.0$ mol-1,
$\eqalign{ & {\Delta _f}H^\circ [(C{O_2})(g)] = - 394.0\,kJ\,mol{ ^{-1}} \cr & S^\circ [Sn{O_2}(s)] = 56.0J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [Sn(s)] = 52.0\,J\,K{ ^{-1}}mo{l^{ - 1}} \cr & S^\circ [C(s)] = 6.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [C{O_2}(g)] = 210.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr} $
Assume that, the enthalpies and the entropies are temperature independent.
Explanation:
$Sn{O_2}(s) + C(s)\buildrel {} \over \longrightarrow Sn(s) + C{O_2}(g)$
Standard enthalpy of reaction,
$\Delta H_{{R^\pi }}^o = (\Delta H_f^oC{O_2}(g)) - (\Delta H_f^oSn{O_2}(s))$
$\Delta H_{{R^\pi }}^o = - 394 - ( - 581) \Rightarrow 187$
Standard entropy of reaction,
$\Delta S_{{R^\pi }}^o = \Delta S_{\Pr oducts}^o - \Delta S_{{\mathop{\rm Re}\nolimits} ac\tan ts}^o$
$\Delta S_{{R^\pi }}^o = [S_{}^o(Sn(s)) + S_{}^o(C{O_2}(g))] - S_{}^o(Sn{O_2}(s)) + S_{}^o(C(s))]$
$\Delta S_{{R^\pi }}^o = [52 + 210] - [56 + 6]$
$ \Rightarrow $ 200 JK$-$1 mol$-$1
We know that, $\Delta H^\circ = T\Delta S^\circ $
$ \therefore $ $T = {{\Delta H^\circ } \over {\Delta S^\circ }}$
$ = {{187 \times 1000} \over {200}} \Rightarrow 935K$
For the reaction to be spontaneous, the temperature should be greater than 935 K.
${P_{H2}}$ is the minimum partial pressure of ${H_2}$ (in bar) needed to prevent the oxidation at $1250$ $K.$ The value of $\ln \left( {{P_{H2}}} \right)$ is ________.
Given: total pressure $=1$ bar, $R$ (universal gas constant ) $=$ $8J{K^{ - 1}}\,\,mo{l^{ - 1}},$ $\ln \left( {10} \right) = 2.3.\,$ $Cu(s)$ and $C{u_2}O\left( s \right)$ are naturally immiscible.
At $1250$ $K:2Cu(s)$ $ + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{O_2}\left( g \right) \to C{u_2}O\left( s \right);$ $\Delta {G^ \circ } = - 78,000J\,mo{l^{ - 1}}$
${H_2}\left( g \right) + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{O_2}\left( g \right) \to {H_2}O\left( g \right);$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\Delta {G^ \circ } = - 1,78,000J\,mo{l^{ - 1}};$ ($G$ is the Gibbs energy)
Explanation:
$2Cu(s) + {1 \over 2}{O_2}(g) \to C{u_2}O(s);\Delta {G^o} = - 78,000$ J mol$-$1 ...... (1)
${H_2}(g) + {1 \over 2}{O_2}(g) \to {H_2}O(g);\Delta {G^o} = - 1,78,000$ J mol$-$1 ..... (2)
Subtracting Eq. (1) $-$ Eq. (2), we get
$2Cu(s) + {H_2}O(g) \to C{u_2}O(s) + {H_2}(g);\Delta {G^o} = + 100000$ J mol$-$1 ..... (3)
Now, $\Delta G = \Delta {G^o} + RT\ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right)$
For the reaction (3) to not occur,
$\Delta G > 0$ or $\Delta {G^o} + RT\ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right) > 0$
$ \Rightarrow 100000 + 8 \times 1250\ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right) > 0$
$ \Rightarrow 100000 + 8 \times 1250\ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right) > {{ - 100000} \over {8 \times 1250}}$
$ \Rightarrow \ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right) > - 10$
$\ln {p_{{H_2}}} > - 10 + \ln {p_{{H_2}O}}$
Now, ${p_{{H_2}O}} = {x_{{H_2}O}} \times {p_T} = 0.01 \times 1 = {10^{ - 2}}$
So, $\ln {p_{{H_2}}} > - 10 - 2\ln 10 \Rightarrow {p_{{H_2}}} > - 14.6$ bar
One mole of an ideal gas is taken from $\mathbf{a}$ to $\mathbf{b}$ along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is $W_{\text {s }}$ and that dotted line path is $W_{\mathrm{d}}$, then the integer closest to the ratio $W_{\mathrm{d}} / W_{\mathrm{s}}$ is
Explanation:
For calculating work done, we need to calculate the area under curve for solid and dotted lines.
Let ' $w_d$ and ' $w$ ' be work done along the dotted and solid path respectively.
$ \begin{aligned} & \mathrm{W}_d=\text { Area ABCD }+ \text { Area EFGC + Area FGIH } \\\\ & w_d =4 \times 1.5+1 \times 1+2.5 \times 2 / 3 \\\\ & =8.65 \end{aligned} $
$ \begin{aligned} &\text { Process of work done }\left(w_s\right) \text { is isothermal }\\\\ &\begin{aligned} w_s & =2 \times 2.303 \log \frac{5.5}{0.5} \\\\ & =2 \times 2.303 \times \log 11 \\\\ & =2 \times 2.303 \times 1.0414=4.79 \\\\ \frac{w_d}{w_s} & =\frac{8.65}{4.79}=1.80 \simeq 2 \end{aligned} \end{aligned} $In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K$^{-1}$, the numerical value for the enthalpy of combustion of the gas in kJ mol$^{-1}$ is ____________.
Explanation:
To find the numerical value for the enthalpy of combustion of the gas in kJ mol$^{-1}$, we first need to determine the total heat released by the combustion of the gas within the calorimeter. We then convert this amount of heat into per mole of the gas. Step 1: Calculate the total heat released, $ q $.
The heat released, $ q $, due to combustion in the calorimeter can be calculated using the formula:
$ q = C \cdot \Delta T $
where:
- $ C $ is the heat capacity of the calorimeter, and
- $ \Delta T $ is the change in temperature.
In this problem:
- $ C = 2.5 \text{ kJ K}^{-1} $
- $ \Delta T = 298.45 \text{ K} - 298.0 \text{ K} = 0.45 \text{ K} $
Substituting these values into the equation gives:
$ q = 2.5 \text{ kJ K}^{-1} \times 0.45 \text{ K} = 1.125 \text{ kJ} $
The total heat released by the process is therefore 1.125 kJ, where this amount of heat is a measure of energy released and absorbed by the calorimeter, therefore it is positive.
Step 2: Convert the heat released to a molar basis.To convert the heat released into per mole of the gas, we first need to calculate the number of moles of the gas that was burnt. The number of moles, $ n $, can be calculated from the mass of the gas and its molecular weight:
$ n = \frac{\text{mass}}{\text{molecular weight}} $
In this problem:
- The mass of the gas = 3.5 g
- Molecular weight of the gas = 28 g mol$^{-1}$
Substituting these values gives:
$ n = \frac{3.5 \text{ g}}{28 \text{ g mol}^{-1}} = 0.125 \text{ mol} $
Step 3: Calculate the enthalpy of combustion per mole.The enthalpy of combustion per mole, $ \Delta H $, is given by:
$ \Delta H = \frac{q}{n} $
Substituting the values we obtained:
$ \Delta H = \frac{1.125 \text{ kJ}}{0.125 \text{ mol}} = 9 \text{ kJ mol}^{-1} $
Therefore, the enthalpy of combustion of the gas is $ -9 \text{ kJ mol}^{-1} $.
Note: The negative sign indicates that the process is exothermic (releases heat).
For the reaction, $2 \mathrm{CO}+\mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2 ; \Delta \mathrm{H}=-560 \mathrm{~kJ}$. Two moles of CO and one mole of $\mathrm{O}_2$ are taken in a container of volume 1 L . They completely form two moles of $\mathrm{CO}_2$, the gases deviate appreciably from ideal behaviour. If the pressure in the vessel changes from 70 to 40 atm , find the magnitude (absolute value) of $\Delta \mathrm{U}$ at 500 K . $(1 \mathrm{~L} \mathrm{~atm}=0.1 \mathrm{~kJ})$
Explanation:
Given,
$ \begin{aligned} \Delta \mathrm{H} & =-560 \mathrm{~kJ} \\ \mathrm{~V} & =1 \mathrm{~L} \\ \mathrm{P}_1 & =70 \mathrm{~atm} \\ \mathrm{P}_2 & =40 \mathrm{~atm} \end{aligned} $
To Find : $\Delta \mathrm{U}$
The $\Delta \mathrm{H}$ is the change in enthalpy, $\Delta \mathrm{U}$ is the change in internal energy, V is the volume of the container, $\mathrm{P}_1$ is the initial pressure and $\mathrm{P}_2$ is the final pressure.
The change in enthalpy is related to change in internal energy as follows:
$ \Delta H=\Delta U+V \Delta P $
As there is no change in the volume, the volume remain constant and $\Delta \mathrm{V}=0$.
$ \Delta \mathrm{H}=\Delta \mathrm{U}+\mathrm{V}\left(\mathrm{P}_2-\mathrm{P}_1\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$
Substituting the respective values in equation (i),
$ \begin{aligned} & -560 \mathrm{~kJ}=\Delta \mathrm{U}+1 \mathrm{~L}(40-70) \mathrm{atm} \\ & -560 \mathrm{~kJ}=\Delta \mathrm{U}+(-30) \mathrm{L} \mathrm{~atm} \end{aligned} $
The $\mathrm{L}-\mathrm{atm}$ values needs to be converted into kJ .
$ \begin{aligned} 1 \mathrm{~L} \mathrm{~atm} & =0.1 \mathrm{~kJ} \\ -560 \mathrm{~kJ} & =\Delta \mathrm{U}+(-30 \times 0.1) \mathrm{kJ} \\ -560 \mathrm{~kJ} & =\Delta \mathrm{U}+(-3) \mathrm{kJ} \\ \therefore \quad \Delta \mathrm{U} & =-560 \mathrm{~kJ}+3 \mathrm{~kJ}=-557 \mathrm{~kJ} \end{aligned} $
Hence, the magnitude (absolute value) of DU is -557 kJ .
$[\mathrm{R}=$ gas constant, $\mathrm{F}=$ Faraday constant, $\mathrm{T}=$ Temperature $]$
(p : pressure, V : volume, T : temperature, H : enthalpy, S : entropy)
$w = - \int {dV{p_{ext}}} $
For a system undergoing a particular process, the work done is
$w = - \int {dV\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)} $
This equation is applicable to a
If ${T_2} > {T_1},$ the correct statement(s) is (are) (Assume $\Delta {H^ \circ }$ and $\Delta {S^ \circ }$ are independent of temperature and ratio of $lnK$ at ${T_1}$ to $lnK$ at ${T_2}$ is greater than ${{{T_2}} \over {{T_1}}}.$ Here $H,$ $S,G$ and $K$ are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)
The correct option(s) is (are)
The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is(are) correct?
For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is(are) correct? (Take $\Delta$S as change in entropy and W as work done)
