Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

The first reaction is exothermic and the second one is endothermic. If a mixture of steam and O₂ is passed over coke and temperature is constant, the conversion of each to CO should not show any heat change, i.e., total heat evolved in I = total heat absorbed in II.

$$\begin{gathered} \therefore n_1\times 2\times 100=n_{2}132 \\ \therefore \frac{n_1}{n_2}=\frac{132}{220}=\frac{0.6}{1} \end{gathered}$$

The statements given under (B), (C) are not correct because statement (B) refers to an endothermic reaction and the standard state for carbon is graphite under (C). For the calculation of $∆H$ of the reaction under (D), additional data in the heat of vapourisation of Br₂(I) is necessary.

$$\begin{gathered} S{i}_2{H}_6\left(g\right) + H_2\left(g\right)\rightleftharpoons 2Si{H}_4\left(g\right) \\ ∆H=-11.7 kJ/mol \\ +80.3 kJ/mol 0 x \\ 2x-80.3=-11.7, \therefore 2x=80.3-11.7=68.6 kJ/mol \\ \therefore x=34.3 kJ/mol \\ Si{H}_4\left(g\right)\to Si{H}_2\left(g\right)+H_2\left(g\right); ∆H=+239.7 kJ/mol \\ +34.3 y 0 \\ y-34.3 =239.7 \\ \therefore y=\left(239.7+34.3\right) kJ mo{l}^{-1}=274 kJ/mol \end{gathered}$$

Pressure of nitrogen in Y = 3.8 atm.

Pressure of nitrogen in X = 1.9 atm.

Pressure of H₂O(g) in X at 300 K = 2-1.9 = 0.1 atm

Pressure of N₂ at 320 K : $\frac{3.8}{300}\times 320$= 4.05 atm.

Total pressure at 320 K : 4.32 atm.

Pressure of water vapour at 320 K = 4.32-4.05 = 0.27 atm.

$$\begin{gathered} \therefore \ln \frac{0.27}{0.1}=\frac{∆H}{R}\left[\frac{1}{300}-\frac{1}{320}\right], \\ \therefore ∆H=39.637 kJ mo{l}^{-1} \end{gathered}$$

According to Kirchhoff's equation,

$$\begin{gathered} ∆H_2=∆H_1+{\int }_{T_1}^{T_2}∆C_P dT \\ =-25000 +{\int }_{300}^{450}\left(40+5\times {10}^{-3} T\right) dT=-18.72 kJ \end{gathered}$$

The equation for combustion of C₂H₄ is

$$\begin{gathered} C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+2H_{2}O\left(l\right) \\ 1 mole 3 moles 2 moles \\ ∆H=∆E+2∆nT=-332190+2\times \left(-2\right)\times \left(273+17\right) \\ =-333350 cals=-333.35 k cals \end{gathered}$$

We have to calculate the enthalpy of the reaction 

OH(g)$\to$O(g) + H(g)

From the given reactions, this can be obtained as follows.

$$\begin{gathered} -\left[\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to OH\left(g\right)\right]; ∆H=-42.09 kJ mo{l}^{-1} \\ +\frac{1}{2}\left[H_2\left(g\right)\to 2H\left(g\right)\right]; ∆H=\frac{1}{2}\times 435.89 kJ mo{l}^{-1} \\ +\frac{1}{2}\left[O_2\left(g\right)\to 2O\left(g\right)\right]; ∆H=\frac{1}{2}\times 495.05 kJ mo{l}^{-1} \\ Add \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ OH\left(g\right)\to H\left(g\right)+O\left(g\right) \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ ∆H=423.38 kJ mol \end{gathered}$$

The given data are

(i) $H_2\left(g\right)\to 2H\left(g\right) ∆H=435 kJ mo{l}^{-1}$

(ii) $C{l}_2\left(g\right)\to 2Cl\left(g\right) ∆H=243 kJ mo{l}^{-1}$

(iii) $HCl\left(g\right)\to H\left(g\right)+Cl\left(g\right) ∆H=431 kJ mo{l}^{-1}$

We have to find $∆H$ for the reaction

$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}C{l}_2\left(g\right)\to HCl\left(g\right)$

This equation can be obtained by the following manipulatipon.

$\frac{1}{2}Eq.\left(i\right)+\frac{1}{2}Eq.\left(ii\right)-Eq. \left(iii\right)$

Hence, carrying out the corresponding manipulation on $∆Hs$, we get

$$\begin{gathered} ∆H=+\frac{1}{2}∆H_{\left(i\right)}+\frac{1}{2}∆H_{\left(ii\right)}-∆H_{\left(iii\right)} \\ =\left(\frac{1}{2}\times 435\frac{1}{2}\times 243-431\right) kJ mo{l}^{-1}=-92 kJ mo{l}^{-1}. \end{gathered}$$

The given data are :

(i) $H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right) ; ∆H^{\circ }=-241 kJ mo{l}^{-1}$

(ii) $$\begin{gathered} C_6{H}_{10}\left(g\right)+\frac{17}{2}{O}_2\left(g\right)\to 6C{O}_2\left(g\right)+5H_{2}O\left(l\right) \\ ∆H^{\circ }=-3800 kJ mo{l}^{-1} \end{gathered}$$

(iii) $$\begin{gathered} C_6{H}_{12}\left(g\right)+9O_2\left(g\right)\to 6C{O}_2\left(g\right)6H_{2}O\left(l\right) \\ ∆H^{\circ }=-3920 kJ mo{l}^{-1} \end{gathered}$$

We have to calculate the enthalpy change for the reaction 

$C_6{H}_{10}\left(g\right)+H_2\left(g\right)\to C_6{H}_{12}\left(g\right)$

This equation can be obtained by the following manipulations.

Eq.(ii) + Eq.(i) - Eq.(iii)

Carrying out the corresponding manipulations on $∆H s,$ we get

$$\begin{gathered} ∆H^{\circ }=∆H_{\left(ii\right)}^{\circ }+∆H_{\left(i\right)}^{\circ }-∆H_{\left(iii\right)}^{\circ } \\ =\left(-3800-241+3920\right) kJ mo{l}^{-1}=-121 kJ mo{l}^{-1}. \end{gathered}$$

The combustion reactions are

$$\begin{gathered} C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+2H_{2}O\left(l\right) \\ C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right) \end{gathered}$$

Let V be the volume of C₂H₄(g) in the gaseous mixture of 3.67 L.n From the chemical equations, we find that

Volume of CO₂(g) produced due to the combustion of C₂H₄(g) = 2V

Volume of CO₂(g) produced due to the combustion of CH₄(g) = 3.67 L - V

Equating the latter with 6.11 L - 2V, we get

3.67 L - V = 6.11 L - 2V or V = 2.44 L

Hence, in the original mixture, we have

Volume of  C₂H₄(g) per litre of the mixture

$=\left(\frac{2.44 L}{3.67 L}\right)\left(1 L\right)=0.665 L$

Volume of CH₄(g) per litre of the mixture

= 1.0 L - 0.665 L = 0.335 L

Now, Volume of 1 mol of any gas at ${25}^{\circ }C$

$=\left(22.414 L\right)\left(\frac{298 K}{273 K}\right)=24.467 L$

Hence, Heat released due to the combustion of C₂H₄(g) 

$=\left(1423 kJ\right)\left(\frac{0.665 L}{24.467 L}\right)=38.68 kJ$

Heat released due to the combustion of CH₄(g) 

$=\left(891 kJ\right)\left(\frac{0.335 L}{24.467 L}\right)=12.20 kJ$

Total heat released = (38.68 + 12.20) kJ = 50.88 kJ.