Q: Derive an expression for work done in adiabatic expansion

Adiabatic process : \$PV^{\gamma} = K\$So, \$P = KV^{-\gamma}\$Work done \$W = \int PdV\$Or \$W = \int KV^{-\gamma} dV\$Or \$W = K \times \dfrac{V^{-\gamma+1}}{1-\gamma}\bigg|_{V_1}^{V_2}\$Or \$W = \dfrac{K}{1-\gamma}\times [{V_2^{-\gamma+1}-V_1^{-\gamma+1}}]\$Or \$W = \dfrac{1}{1-\gamma}\times [{KV_2^{-\gamma+1}-KV_1^{-\gamma+1}}]\$Or \$W = \dfrac{1}{1-\gamma}\times [P_2V_2^{\gamma}{V_2^{-\gamma+1}-P_1V_1^{\gamma}V_1^{-\gamma+1}}]\$Or \$W = \dfrac{P_2V_2-P_1V_1}{1-\gamma}\$

Q: Calculate the standard enthalpy of formation of \$CH_3OH(I)\$ from the following data :\$CH_3OH (I) +\dfrac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l); \, \triangle_rH^{\theta}=-726 kJ mpl^{-1}\$\$C_{(graphite)}+O_2(g) \rightarrow CO_2(g); \triangle_cH^{\theta}=- 393 kJ mol^{-1}\$\$H_2(g)+\dfrac{1}{2}O_2(g) \rightarrow H_2O(l); \triangle_fH^{\theta}=-286 kJ mol^{-1}\$

On reversing the first equation we get, \$CO_2(g) + 2H_2O(l) \rightarrow CH_3OH(l) + \dfrac{3}{2}O_2(g)\$; \$\Delta{_rH^{\theta}} = +726kJmol^{-1}\$ (on reversing the reaction, the sign of \$\Delta{H}\$ also reverses)The second equation remains as such,\$C_{(graphite)} + O_2(g) \rightarrow CO_2(g); \Delta{_cH^{\theta}} = -393kJmol^{-1}\$On multiplying equation three by 2 we get,\$2H_2(g) + O_2(g) \rightarrow 2H_2O(l); 2\Delta{_fH^{\theta}} = 2 \times -286kJmol^{-1} = -572kJmol^{-1}\$On adding these three equations we get,\$C_{(graphite)} + 2H_2(g) + \dfrac{1}{2}O_2(g) \rightarrow CH_3OH(l)\$Enthalpy of formation of \$CH_3OH = \Delta{_fH} = \Delta{_rH^{\theta}} + \Delta{_cH^{\theta}} + 2\Delta{_fH^{\theta}}\$\$\Delta{_fH} = +726 - 393 - 572 kJmol^{-1} = -239 kJmol^{-1}\$

Q: The equilibrium constant for a reaction is 10. what will be value of \$ \displaystyle \bigtriangleup G^{\ominus } \$ ? R =8.314 \$ \displaystyle JK^{-1} mol^{-1} \$, T=300 K

\$\displaystyle \Delta G^0 = -2.303 RT log K = -2.303 \times 8.314 \times 300log10 = -5.744 kJ/mol\$

Q: For the reaction, \$2Cl(g) \rightarrow Cl_{2}(g)\$, the signs of \$\Delta H\$ and \$\Delta S\$ respectively, are

\$2Cl(g) \rightarrow Cl_{2} (g)\$Entropy is decreasing \$(-ve)\$ in the reaction as no. of molecules are decreasing which reduces the randomness of the system. Further the reaction is exothermic since a bond is being formed, i.e., \$\triangle H\$ is also \$-ve\$.Hence, option C is correct

Q: For the reaction \$NH_4Cl(S) \rightarrow NH_3(g) + HCl(g)\$ at \$25^0\$C, enthalpy change \$\triangle = +177 kJ mol^{-1}\$ and entropy change \$\triangle S = +285 JK^{-1}mol^{-1}\$. Calculate free energy change \$\triangle G\$ at \$25^0\$C and product whether the reaction is spontaneous or not.

Given, \$\triangle H= 177 \ kJ/mol, \ \triangle S=285 \ JK^{-1}mol^{-1}, \ T=298 \ K\$Using the relation, \$\triangle G=\triangle H- T\triangle S\$\$\triangle G= (177 \times 10^{3} \ J/mol)- (298 \ K) (285 \ JK^{-1}mol^{-1})\$\$\triangle G= 92070 \ J/mole\$\$\boxed {\triangle G= 92.07 \ kJ/mole}\$\$\boxed {\triangle G >0}\$Since \$\triangle G\$ is positive, Hence reaction is non spontaneous.

Question 1

Derive an expression for work done in adiabatic expansion

Accepted Answer

Adiabatic process :    $PV^{\gamma} = K$So,     $P = KV^{-\gamma}$Work done    $W = \int PdV$Or     $W = \int KV^{-\gamma} dV$Or     $W = K \times \dfrac{V^{-\gamma+1}}{1-\gamma}\bigg|_{V_1}^{V_2}$Or     $W = \dfrac{K}{1-\gamma}\times [{V_2^{-\gamma+1}-V_1^{-\gamma+1}}]$Or     $W = \dfrac{1}{1-\gamma}\times [{KV_2^{-\gamma+1}-KV_1^{-\gamma+1}}]$Or     $W = \dfrac{1}{1-\gamma}\times [P_2V_2^{\gamma}{V_2^{-\gamma+1}-P_1V_1^{\gamma}V_1^{-\gamma+1}}]$Or     $W = \dfrac{P_2V_2-P_1V_1}{1-\gamma}$

Question 2

Calculate the standard enthalpy of formation of $CH_3OH(I)$ from the following data :$CH_3OH (I) +\dfrac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l); \, \triangle_rH^{\theta}=-726 kJ  mpl^{-1}$$C_{(graphite)}+O_2(g) \rightarrow CO_2(g); \triangle_cH^{\theta}=- 393 kJ mol^{-1}$$H_2(g)+\dfrac{1}{2}O_2(g) \rightarrow H_2O(l); \triangle_fH^{\theta}=-286 kJ mol^{-1}$

Accepted Answer

On reversing the first equation we get, $CO_2(g) + 2H_2O(l) \rightarrow CH_3OH(l) + \dfrac{3}{2}O_2(g)$; $\Delta{_rH^{\theta}} = +726kJmol^{-1}$ (on reversing the reaction, the sign of $\Delta{H}$ also reverses)The second equation remains as such,$C_{(graphite)} + O_2(g) \rightarrow CO_2(g); \Delta{_cH^{\theta}} = -393kJmol^{-1}$On multiplying equation three by 2 we get,$2H_2(g) + O_2(g) \rightarrow 2H_2O(l); 2\Delta{_fH^{\theta}} = 2 \times -286kJmol^{-1} = -572kJmol^{-1}$On adding these three equations we get,$C_{(graphite)} + 2H_2(g) + \dfrac{1}{2}O_2(g) \rightarrow CH_3OH(l)$Enthalpy of formation of $CH_3OH = \Delta{_fH} = \Delta{_rH^{\theta}} + \Delta{_cH^{\theta}} + 2\Delta{_fH^{\theta}}$$\Delta{_fH} = +726 - 393 - 572 kJmol^{-1} = -239 kJmol^{-1}$

Question 3

Which of the following statement is false?

Accepted Answer

Work is not a state function. It is path function. Its value depends on the path chosen to reach a particular state.

Question 4

State Hess's law of constant heat summation and explain it with an example.

Accepted Answer

Hess's law of constant heat summation:                         The law states that the change&#10; in enthalpy for a reaction is the same whether the reaction takes place&#10; in one or a series of steps. The Hess's law can also be stated as the &#10;enthalpy change for a chemical reaction is the same regardless of the &#10;path by which the reaction occurs.For example, consider following two paths for the preparation of methylene chloridePath I : $\displaystyle CH_4(g) + 2Cl_2(g) \rightarrow CH_2Cl_2(g) + 2HCl (g) \: \: \: \Delta H^0_1=-202.3 \: kJ $Path II : $\displaystyle CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl (g) \: \: \: \Delta H^0_2=-98.3 \: kJ $$\displaystyle CH_3Cl(g) + Cl_2(g) \rightarrow CH_2Cl_2(g) + HCl (g) \: \: \: \Delta H^0_3=-104.0 \: kJ $Adding two steps$\displaystyle CH_4(g) + 2Cl_2(g) \rightarrow CH_2Cl_2(g) + 2HCl (g) \: \: \: \Delta H^0=-202.3 \: kJ $Thus whether we follow path I or path II, the enthalpy change of the reaction is same. $\displaystyle \Delta H^0_1=\Delta H^0_2+\Delta H^0_3=-202.3 \: kJ $

Question 5

Which of the following statements is incorrect?

Accepted Answer

Critical temperature of the substance is the temperature at a substance's critical point.Above $T_c$ we cannot liquify gas keep pressure constant although on $\uparrow$ pressure we can liquify.

Question 6

The equilibrium constant for a reaction is 10. what will be value of $ \displaystyle \bigtriangleup G^{\ominus } $ ? R =8.314 $ \displaystyle JK^{-1}  mol^{-1}  $, T=300 K

Accepted Answer

$\displaystyle \Delta G^0 = -2.303 RT log K = -2.303 \times 8.314 \times 300log10 = -5.744 kJ/mol$

Question 7

For the reaction, $2Cl(g) \rightarrow Cl_{2}(g)$, the signs of $\Delta H$ and $\Delta S$ respectively, are

Accepted Answer

$2Cl(g) \rightarrow Cl_{2} (g)$Entropy is decreasing $(-ve)$ in the reaction as no. of molecules are decreasing which reduces the randomness of the system. Further the reaction is exothermic since a bond is being formed, i.e., $\triangle H$ is also $-ve$.Hence, option C is correct

Question 8

Which of the following is an extensive property ?

Accepted Answer

A property of a system, whose magnitude depends upon the amount of matter, is known as extensive property. Properties like mass, volume, internal energy, heat content, free energy, enthalpy, entropy, heat capacity, surface area, energy, etc. are dependent upon the mass of the system and are called extensive properties.Intensive property: A property of a system, whose magnitude does not depend upon the amount of matter, is known as intensive property. Properties like temperature, pressure, surface tension, viscosity, specific heat, molar energy, molar entropy, density, refractive index, etc., are independent of the mass of the system and are called intensive properties.

Question 9

Predict in which of the following entropy increases/ decreases.(a) A liquid crystallizes into a solid.(b) Temperature f a crystalline solid is raised from 0 K to 115 K.(c) $2Na{ HCO }_{ 3 }\left( s \right) \longrightarrow Na{ NO }_{ 3 }\left( s \right) +{ CO }_{ 2 }\left( g \right) +{ H }_{ 2 }O\left( g \right) $(d) ${ H }_{ 2 }\left( g \right) \longrightarrow 2H\left( g \right) $

Accepted Answer

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Question 10

For the reaction $NH_4Cl(S) \rightarrow NH_3(g) + HCl(g)$ at $25^0$C, enthalpy change $\triangle = +177 kJ mol^{-1}$ and entropy change $\triangle S = +285 JK^{-1}mol^{-1}$. Calculate free energy change $\triangle G$ at $25^0$C and product whether the reaction is spontaneous or not.

Accepted Answer

Given, $\triangle H= 177 \ kJ/mol, \ \triangle S=285 \ JK^{-1}mol^{-1}, \ T=298 \ K$Using the relation, $\triangle G=\triangle H- T\triangle S$$\triangle G= (177 \times 10^{3} \ J/mol)- (298 \ K) (285 \ JK^{-1}mol^{-1})$$\triangle G= 92070 \ J/mole$$\boxed {\triangle G= 92.07 \ kJ/mole}$$\boxed {\triangle G >0}$Since $\triangle G$ is positive, Hence reaction is non spontaneous.