Chemistry MCQs for NEET — Practice Questions with Answers

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.

The combustion of cyclopropane involves the formation of CO2 and H2O from the cyclopropane molecule. The enthalpy change can be calculated using the given enthalpies of formation and the enthalpy of isomerization of cyclopropane to propene, which acts as an intermediate step.

H3BO3 is a Lewis acid that can accept an electron pair to form a conjugate base. The removal of a proton from H3BO3 leads to the formation of the conjugate base H2BO3-, which is a tetrahedral oxoanion containing a boron atom with a formal negative charge.

(b)  PCl₅ (g) $\rightleftharpoons$ PCl₃ (g) + Cl₂ (g)

         1               0              0

        1-$\mathrm{\alpha }$             $\mathrm{\alpha }$              $\mathrm{\alpha }$ 

.^..   K_p = $\frac{{\mathrm{\alpha }}^2}{(1-\mathrm{\alpha })}\left[\frac{\mathrm{P}}{1+\mathrm{\alpha }}\right]=\frac{{\mathrm{\alpha }}^2\mathrm{P}}{1-{\mathrm{\alpha }}^2}$

         or $\alpha =\sqrt{\frac{K_p}{P}}$ if 1-${\mathrm{\alpha }}^2$=1