Enthalpy MCQs

Ruhul Amin Alig

Question: From NCERT NEET 2020

For the reaction, 2Cl(g) $\to$ Cl2(g), the correct option is:

1. $∆_{r} H > 0 and ∆_{r} S < 0$

2. $∆_{r} H < 0 and ∆_{r} S > 0$

3. $∆_{r} H < 0 and ∆_{r} S < 0$

4. $∆_{r} H > 0 and ∆_{r} S > 0$

Answer ▽

Correct option is

$Δ_{r} H < 0$ and $Δ_{r} S < 0$

$2 C l (g) \to C l_{2} (g)$
Atoms forming a molecule, making system less disordered. Entropy is decreasing $Δ S (- v e)$ in the reaction. Further the addition reaction is exothermic since a bond is being formed, therefore $Δ H$ is also $- v e .$

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Question: From NCERT NEET 2012

Standard enthalpy of vaporisation $∆$ vapH $°$ for water at 100 $°$ C is 40.66 kJ mol-1. The

internal energy of vaporisation of water at 100 $°$ C (in kJ mol-1) is-

(Assume water vapour to behave like an ideal gas)

1. +37.56

2. -43.76

3. +43.76

4. +40.66

Answer ▽

Correct option is

$37.53$ KJ $m o l^{- 1}$

Solution:- (A) $37.53 K J / m o l$
We know that for gaseous reactants and products , we have a relation between standard enthalpy of vapourization $(Δ H_{v a p .})$ and standard internal energy $(Δ E)$ as-
$Δ H_{v a p .} = Δ E + Δ n_{g} R T$
whereas,
$Δ n_{g} = n_{2} - n_{1}$ , i.e., difference between no. of moles of reactant and product.
For vapourization of water,
$H_{2} O_{(l)} ⟶ H_{2} O_{(g)}$
$∴ Δ n_{g} = 1 - 0 = 1$
$T = 100 ℃ = (100 + 273) = 373 K$
$∴ Δ H_{v a p .} = Δ E + Δ n_{g} R T$
$\Rightarrow Δ E = Δ H_{v a p .} - Δ n_{g} R T = 40.63 - (1 \times 8.314 \times 10^{- 3} \times 373) = 37.53 K J / m o l$
Hence the value $Δ E$ for this process will be $37.53 K J / m o l$ .

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Question: From NCERT NEET 2011

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol-1 at 27°C,

the entropy change for the process would be

1. 1.0 J mol-1 K-1

2. 0.1 J mol-1K-1

3. 100 J mol-1K-1

4. 10 J mol-1K-1

Answer ▽

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Question: From NCERT NEET 2011

The given reaction is

$4 H (g) \to 2 H_{2} (g) is - 869.6 kJ$

The enthalpy change for the reaction is -869.6 kJ. The dissociation energy of the H-H bond is -

1. -869.6 kJ

2. +434.8kJ

3. +217.4kJ

4. -434.8 kJ

Answer ▽

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Question: From NCERT NEET 2011

The value of $∆ H$ for the reaction is less than zero. Formation of $X_{2} (g) + 4 Y_{2} (g) ⇋ 2 X Y_{4} (g)$ will be favoured at

1. low pressure and low temperature

2. high temperature and low pressure

3. high pressure and low temperature

4. high temperature and high pressure

Answer ▽

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Question: From NCERT NEET 2007

Consider the following reactions :

(i) $H^{+} (aq) + {OH}^{-} (aq) = H_{2} O (l) ∆ H = - x_{1} kJ {mol}^{- 1}$

(ii) $H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (l) ∆ H = - x_{2} kJ {mol}^{- 1}$

(iii) ${CO}_{2} (g) + H_{2} (g) = CO (g) + H_{2} O (l) ∆ H = - x_{3} kJ {mol}^{- 1}$

(iv) $C_{2} H_{5} (g) + \frac{5}{2} O_{2} (g) = 2 {CO}_{2} (g) + H_{2} O (l) ∆ H = - x_{4} kJ {mol}^{- 1}$

Enthalpy of formation of H2O(l) is:

1. -x2 kJ mol-1

2. +x3 kJ mol-1

3. -x4 kJ mol-1

4. -x1 kJ mol-1

Answer ▽

The amount of heat evolved or absorbed during the formation of 1 mole of a compound from its constituent is known as heat of formation. So, the correct answer is

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Question: From NCERT NEET 2007

Given those bond energies of H-H and Cl-Cl are 430 kJ mol-1 and 240 kJ mol-1 respectively and ΔHf for HCI is -90 kJ mol-1. Bond enthalpy of HCl is:

1. 290 kJ mol-1

2. 380 kJ mol-1

3. 425 kJ mol-1

4. 245 kJ mol-1

Answer ▽

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Question: From NCERT Derived PYQ

Change in enthalpy for reaction,

2H2O2(l) $\to$ 2H2O(l) + O2(g)

If the heat of formation of H2O2(l) and H2O(l) are -188 and -286 kJ/mol respectively is -

1. -196 kJ/mol

2. + 196 kJ/mol

3. +948 kJ/mol

4. -948 kJ/mol

Medium

Answer ▽

1
For $2$ moles $H_{2} O_{2} = - 2 \times 188 k J / m o l$

$△ H_{f} [H_{2} O (l)] = - 286 k J / m o l$

For $2$ moles $H_{2} O = - 2 \times 286 k J / m o l$

$△ H_{r} = (2 \times - 286) - (2 \times - 188)$

$= - 196 k J / m o l$ .

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Question: From NCERT Derived PYQ

Enthalpy of the reaction,

CH4(g) + 1/2 O2(g) $\to$ CH3OH (l), is negative. If the enthalpy of combustion of CH4 and CH3OH are x and y respectively, then which relation is correct?

1. x>y

2. x<y

3. x=y

4. x $⩾$ y

Answer ▽

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Question: From NCERT Derived PYQ

The reaction CH4(g) + Cl2(g) $\to$ CH3Cl(g) + HCl(g)

has $∆ H$ =-25 K cal

From the given data, what is the bond enthalpy of Cl-Cl bond?

1. 70 Kcal

2. 80 Kcal

3. 67.75 Kcal

4. 57.85 Kcal

Answer ▽

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Question: From NCERT Derived PYQ

Consider the reaction

N2(g) + 3H2(g) $\to$ 2NH3(g) carried out at constant temperature and pressure. If $∆$ H and $∆$ E are enthalpy change and internal energy change respectively, which of the following expression is True?

$1 . ∆ H ⩾ ∆ E 2 . ∆ H = ∆ E 3 . ∆ H < ∆ E 4 . ∆ H > ∆ E$

Answer ▽

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Question: From NCERT Derived PYQ

In the reaction at 300 K

H2(g) + Cl2(g) $\to$ 2HCl(g) $∆ H °$ = -185 kJ

if 2 mol of H2 completely react with 2 mole of Cl2 to form HCl. What is $∆ U °$ for this reaction?

(1) 0

(2) -185 kJ

(3) -370 kJ

(4) None of these

Answer ▽

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Question: From NCERT Derived PYQ

What will be the standard enthalpy of reaction for the following reaction using the listed enthalpies of reaction:-

3Co(s) + 2O2(g) $\to$ Co3O4(s); $∆ H$ =?

2Co(s) + O2(g) $\to$ 2CoO(s) $∆ H_{1}$ = -475.8 KJ

6CoO(s) + O2(g) $\to$ 2Co3O4(s) $∆ H_{2}$ = -355.0 KJ

(1) -891.2 KJ

(2) -120.8 KJ

(3) 891.2 KJ

(4) -830.8 KJ

Answer ▽

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Question: From NCERT Derived PYQ

For given following equations and $∆ H °$ values, determine the enthalpy of reaction at 298 K for the reaction

C2H4(g) + 6F2(g) $\to$ 2CF4(g) + 4HF(g)

H2(g) + F2(g) $\to$ 2HF(g) $∆ H_{1}^{°}$ = -537 kJ

C(s) + 2F2(g) $\to$ CF4(g) $∆ H_{2}^{°}$ =-680 kJ

2C(s) + 2H2(g) $\to$ C2H4(g) $∆ H_{3}^{°}$ = 52 kJ

1. -1165

2. -2486

3. +1165

4. +2486

Answer ▽

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Question: From NCERT Derived PYQ

If the bond energies of H-H, Br-Br, and H-Br are 433, 192 and 364 kJ mol-1 respectively,

the $∆ H °$ for the reaction

H2(g) + Br2(g) $\to$ 2HBr(g) is

1. -103 kJ

2. -261 kJ

3. +103 kJ

4. +261 kJ

Answer ▽

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Question: From NCERT Derived PYQ

Enthalpy for the reaction C + O2 → CO2 is

(1) Positive

(2) Negative

(3) Zero

(4) None

Answer ▽

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Question: From NCERT Derived PYQ

If 1.00 kcal of heat is added to 1.2 L of oxygen in a cylinder at constant pressure of 1.000 atm, the volume increases to 1.5 L, Hence $∆ E$ for this process is:

1. 0.993 kcal

2. 1.0073 kcal

3. 0.0993 kcal

4. 1.00073 kcal

Answer ▽

As we know that,

$Δ U = q - P Δ V$

Here, $q = 1 kcal = 1000 cal.$

$P = Pressure = 1 atm$

$Δ V =$ Change in volume $= (1.5 - 1.2) L = 0.3 L$

$∴ Δ U = 1000 - 1 \times 0.3 = 992.7 cal. = 0.993 kcal$

At constant pressure,

$Δ H = q_{P} = 1 kcal$

Hence, $Δ U = 0.993 kcal a n d Δ H = 1 kcal$ .

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Question: From NCERT Derived PYQ

Enthalpy change when 1.00 g water is frozen at $0^{\circ} C,$ is :

$(∆ H_{f u s} = 1.435 k c a l m o l^{- 1})$

1. 0.0797 kcal

2. -0.0797 kcal

3. 1.435 kcal

4. -1.435 kcal

Answer ▽

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Question: From NCERT Derived PYQ

44.0 kJ of heat is required to evaporate one mole of water at 298 K. If $∆ H_{f}$ of $H_{2} O (l)$ is -286 kJ mol-1, $∆ {H^{\circ}}_{f}$ of $H_{2} O (g)$ is

1. -330 kJ mol-1

2. +242 kJ mol-1

3. -242 kJ mol-1

4. -198 kJ mol-1

Answer ▽

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Question: From NCERT NEET Derived PYQ

The relation between ΔE and ΔH is [MP PET 1992; MP PMT 1996; MP PET/PMT 1998]

(1) $ΔH = ΔE - P ΔV$

(2) $ΔH = ΔE + P ΔV$

(3) $ΔE = ΔV + ΔH$

(4) $ΔE = ΔH + P ΔV$

Answer ▽

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Question: From NCERT Derived PYQ

The molar heat capacity, $C_{v}$ of an ideal gas whose energy is that of translational motion only is

1. $2.98 J d e g^{- 1} m o l^{- 1}$

2. $12.47 J d e g^{- 1} m o l^{- 1}$

3. $6.43 J d e g^{- 1} m o l^{- 1}$

4. $9.41 J d e g^{- 1} m o l^{- 1}$

Answer ▽

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Enthalpy MCQs

$Δ_{r} H < 0$ and $Δ_{r} S < 0$

$2 C l (g) \to C l_{2} (g)$
Atoms forming a molecule, making system less disordered. Entropy is decreasing $Δ S (- v e)$ in the reaction. Further the addition reaction is exothermic since a bond is being formed, therefore $Δ H$ is also $- v e .$

$37.53$ KJ $m o l^{- 1}$

1
For $2$ moles $H_{2} O_{2} = - 2 \times 188 k J / m o l$

$△ H_{f} [H_{2} O (l)] = - 286 k J / m o l$

For $2$ moles $H_{2} O = - 2 \times 286 k J / m o l$

$△ H_{r} = (2 \times - 286) - (2 \times - 188)$

$= - 196 k J / m o l$ .

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Enthalpy MCQs

Δr​H<0 and Δr​S<0

2Cl(g)→Cl2​(g) Atoms forming a molecule, making system less disordered. Entropy is decreasing ΔS(−ve) in the reaction. Further the addition reaction is exothermic since a bond is being formed, therefore ΔH is also −ve.

37.53 KJ mol−1

1For 2 moles H2​O2​=−2×188 kJ/mol△Hf​ [H2​O(l)]=−286 kJ/molFor 2 moles H2​O=−2×286 kJ/mol△Hr​=(2×−286)−(2×−188)=−196 kJ/mol.

As we know that,ΔU=q−PΔVHere, q=1 kcal=1000 cal.P=Pressure=1 atmΔV= Change in volume =(1.5−1.2)L=0.3L∴ΔU=1000−1×0.3=992.7cal.=0.993 kcalAt constant pressure,ΔH=qP​=1 kcalHence, ΔU=0.993 kcal andΔH=1 kcal.

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$Δ_{r} H < 0$ and $Δ_{r} S < 0$

$2 C l (g) \to C l_{2} (g)$
Atoms forming a molecule, making system less disordered. Entropy is decreasing $Δ S (- v e)$ in the reaction. Further the addition reaction is exothermic since a bond is being formed, therefore $Δ H$ is also $- v e .$

$37.53$ KJ $m o l^{- 1}$

1
For $2$ moles $H_{2} O_{2} = - 2 \times 188 k J / m o l$

$△ H_{f} [H_{2} O (l)] = - 286 k J / m o l$

For $2$ moles $H_{2} O = - 2 \times 286 k J / m o l$

$△ H_{r} = (2 \times - 286) - (2 \times - 188)$

$= - 196 k J / m o l$ .