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NEET NEET Questions — Free MCQ Practice

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The heat of combustion of ethylene at $17^{\circ} C$ and at constant volume is -332.19 kcals. What is the value at constant pressure, given that water is in liquid state ?

A.

-131.25 k cals

B.

-412.23 k cals

C.

-534.12 k cals

D.

-333.35 k cals

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Explanation

The equation for combustion of C₂H₄ is

$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l) 1 m o l e 3 m o l e s 2 m o l e s ∆ H = ∆ E + 2 ∆ n T = - 332190 + 2 \times (- 2) \times (273 + 17) = - 333350 c a l s = - 333.35 k c a l s$

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The enthalpies of the following reactions are shown alongwith.

$\frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) \to O H (g); ∆ H = 42.09 k J m o l^{- 1} H_{2} (g) \to 2 H (g); ∆ H = 435.89 k J m o l^{- 1} O_{2} (g) \to 2 O (g); ∆ H = 495.05 k J m o l^{- 1}$

Calculate the O-H bond energies for the hydroxyl radical.

A.

223.18 kJ mol^-1

B.

423.38 kJ mol^-1

C.

513.28 kJ mol^-1

D.

113.38 kJ mol^-1

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Explanation

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.

$- [\frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) \to O H (g)]; ∆ H = - 42.09 k J m o l^{- 1} + \frac{1}{2} [H_{2} (g) \to 2 H (g)]; ∆ H = \frac{1}{2} \times 435.89 k J m o l^{- 1} + \frac{1}{2} [O_{2} (g) \to 2 O (g)]; ∆ H = \frac{1}{2} \times 495.05 k J m o l^{- 1} A d d___________________O H (g) \to H (g) + O (g)___________________∆ H = 423.38 k J m o l$

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The bond dissociation enthalpy of gaseous H₂, Cl₂ and HCl are 435, 243 and 431 kJ mol^-1, respectively. Calculate the enthalpy of formation of HCl gas.

A.

-92 kJ mol^-1

B.

-82 kJ mol^-1

C.

-21 kJ mol^-1

D.

-55 kJ mol^-1

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Explanation

The given data are

(i) $H_{2} (g) \to 2 H (g) ∆ H = 435 k J m o l^{- 1}$

(ii) $C l_{2} (g) \to 2 C l (g) ∆ H = 243 k J m o l^{- 1}$

(iii) $H C l (g) \to H (g) + C l (g) ∆ H = 431 k J m o l^{- 1}$

We have to find $∆ H$ for the reaction

$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g)$

This equation can be obtained by the following manipulatipon.

$\frac{1}{2} E q . (i) + \frac{1}{2} E q . (i i) - E q . (i i i)$

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

$∆ H = + \frac{1}{2} ∆ H_{(i)} + \frac{1}{2} ∆ H_{(i i)} - ∆ H_{(i i i)} = (\frac{1}{2} \times 435 \frac{1}{2} \times 243 - 431) k J m o l^{- 1} = - 92 k J m o l^{- 1} .$

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The standard enthalpy of combustion at $25^{\circ} C$ of hydrogen, cyclohexene (C₆H₁₀) and cyclohexane (C₆H₁₂) are -241, -3800 and -3920 kJ mol^-1, respectively. Calculate the standard enthalpy of hydrogenation of cyclohexene.

A.

-131 kJ mol^-1

B.

-155 kJ mol^-1

C.

-167 kJ mol^-1

D.

-121 kJ mol^-1

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Explanation

The given data are :

(i) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); ∆ H^{\circ} = - 241 k J m o l^{- 1}$

(ii) $C_{6} H_{10} (g) + \frac{17}{2} O_{2} (g) \to 6 C O_{2} (g) + 5 H_{2} O (l) ∆ H^{\circ} = - 3800 k J m o l^{- 1}$

(iii) $C_{6} H_{12} (g) + 9 O_{2} (g) \to 6 C O_{2} (g) 6 H_{2} O (l) ∆ H^{\circ} = - 3920 k J m o l^{- 1}$

We have to calculate the enthalpy change for the reaction

$C_{6} H_{10} (g) + H_{2} (g) \to C_{6} H_{12} (g)$

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

$∆ H^{\circ} = ∆ H_{(i i)}^{\circ} + ∆ H_{(i)}^{\circ} - ∆ H_{(i i i)}^{\circ} = (- 3800 - 241 + 3920) k J m o l^{- 1} = - 121 k J m o l^{- 1} .$

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A gas mixture consisting of 3.67 litres of ethylene and methane on complete combustion at $25^{\circ} C$ produces 6.11 litres of CO₂. Find out the amount of heat evolved on burning one litre of the gas mixture. The heats of combustion of ethylene and methane are -1423 and -891 kJ mol^-1, respectively, at $25^{\circ} C$ .

A.

30.88 kJ

B.

20.28 kJ

C.

50.88 kJ

D.

60.18 kJ

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Explanation

The combustion reactions are

$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l) C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$

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

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

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 $25^{\circ} C$

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

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

$= (1423 k J) (\frac{0.665 L}{24.467 L}) = 38.68 k J$

Heat released due to the combustion of CH₄(g)

$= (891 k J) (\frac{0.335 L}{24.467 L}) = 12.20 k J$

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

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From the following data, calculate the enthalpy change for the combustion of cyclopropane at 298 K. The enthalpy of formation of CO₂(g), H₂O(l) and propene (g) are -393.5, -285.8 and 20.42 kJ mol^-1 respectively. The enthalpy of isomerisation of cyclopropane to propene is -33.0 kJ mol^-1.

A.

-1021.32 kJ mol^-1

B.

-2091.32 kJ mol^-1

C.

-5021.32 kJ mol^-1

D.

-3141.32 kJ mol^-1

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Explanation

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.

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The conjugate base of H₃BO₃ is:

A.

$B (OH)_{4}^{-}$

B.

$H_{2} {BO}_{3}^{-}$

C.

${HBO}_{3}^{2 -}$

D.

$H_{4} {BO}_{3}^{+}$

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Explanation

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.

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The degree of dissociation of PCl₅ ( $α$ ) obeying the equilibrium,

PCl₅ (g) $⇌$ PCl₃ (g) + Cl₂ (g), is approximately related to the pressure at equilibrium by:

A.

$α \propto P$

B.

$α \propto 1 / \sqrt{P}$

C.

$α \propto 1 / P^{2}$

D.

$α \propto 1 / P^{4}$

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Explanation

(b) PCl₅ (g) $⇌$ PCl₃ (g) + Cl₂ (g)

1 0 0

1- $α$ $α$ $α$

.^.. K_p = $\frac{α^{2}}{(1 - α)} [\frac{P}{1 + α}] = \frac{α^{2} P}{1 - α^{2}}$

or $α = \sqrt{\frac{K_{p}}{P}}$ if 1- $α^{2}$ =1

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The equilibrium constants $K_{P_{1}}$ and $K_{P_{2}}$ for the reactions X $⇌$ 2Y and Z $⇌$ P + Q respectively are in the ratio of 1:9. If the degree of dissociation ox X and Z be equal then the ratio of total pressure at these equilibria is:

A.

1:9

B.

1:36

C.

1:1

D.

1:3

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Explanation

(b) Given : $\frac{K_{P_{1}}}{K_{P_{2}}}$ = 1/9 and $α_{x} = α_{z}$

$X ⇌ 2 Y 1 0 1 - α 2 α$ .^.. $K_{P_{1}} = \frac{(2 α_{x})^{2}}{(1 - α_{x})} x {[\frac{P_{1}}{1 + α_{x}}]}^{1}$

$Z ⇌ P + Q 1 0 1 - α 2 α$ .^.. $K_{P_{2}} = \frac{(α_{z})^{2}}{(1 - α_{z})} x {[\frac{P_{2}}{1 + α_{z}}]}^{1}$

.^.. P₁/P₂ = $\frac{K_{P_{1}}}{K_{P_{2}}}$ x 1/4 = 1/9 x 1/4 = 1/36

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In a system: $A (s) \overset{}{⇌} 2 B (g) + 3 C (g) .$ If the concentration of C at eqilibrium is increased by a factor 2, it will cause the eqilibrium concentration of B to change to:

A.

two times of its original value

B.

one half of its original value

C.

2 $\sqrt{2}$ times of its original value

D.

$\frac{1}{2 \sqrt{2}}$ times of its original value

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Explanation

For $A (s) \overset{}{⇌} 2 B (g) + 3 C (g) .$

$∴ K_{c} = {[C]}^{3} {[B]}^{2}; if [C] becomes twice, Then let conc . of B becomes B' then K_{c} = {[2 C]}^{3} {[B']}^{2} or {[C]}^{3} . {[B]}^{2} = {[2 C]}^{3} {[B']}^{2} ∴ \frac{[B']}{[B]} = \frac{1}{8} = \frac{1}{2 \sqrt{2}}$

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