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Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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For which change $∆ H \neq ∆ E$

$H_{2} + I_{2} ⇌ 2 H I$

$H C l + N a O H \to N a C l + H_{2} O$

$C (s) + O_{2} (g) ⇌ C O_{2} (g)$

$N_{2} + 3 H_{2} \to 2 N H_{3}$

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Explanation

$∆ H = ∆ E + ∆ n R T$

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

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

0.0797 kcal

-0.0797 kcal

1.435 kcal

-1.435 kcal

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Explanation

$∆ H$ (per g) = $\frac{- 1.435}{18}$ kcal = - 0.0797 kcal g

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One mole of hydrogen gas at $25^{\circ} C$ and 1 atm pressure is heated at constant pressure until its volume has doubled. Given that $C_{v}$ for hydrogen is 3.0 cal deg^-1 mol^-1, the $∆ H$ and $∆ E$ for this process are-

$∆ H$ = 1490 cal and $∆ E$ = 894 cal

$∆ H$ = $∆ E$ = 1490 cal

$∆ H$ = 894 cal and $∆ E$ = 1490 cal

$∆ H$ = $∆ E$ = 894 cal

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Explanation

$∆ H = C_{P} ∆ T a n d ∆ E = C_{V} ∆ T .$

$C_{P} - C_{V} = R$ using these equations after calculating the final temperature, the desired result is obtained.

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The enthalpy change of the reaction

$A l_{2} C l_{6} (s) + 6 N a (s) \to 2 A l (s) + 6 N a C l (s)$ is -257 kcal.

Given the heat of formation of NaCl(s) is 98.0 kcal, the heat of formation of $A l_{2} C l_{6} (s)$ is-

158 kcal

-331 kcal

-166 kcal

316 kcal

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Explanation

$∆ H$ of reaction = $H_{N a C l} \times 6 - H_{A l_{2} C l_{6}} w h e r e H_{N a C l} a n d H_{A l_{2} C l_{6}}$ are the heats of formation of NaCl and Al₂Cl₆.

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Some of the thermodynamic parameters are state variables while some are process variables. Some grouping of the parameters are given. Choose the correct one-

State variables : Temperature, No. of moles

Process variables : Internal energy, work done by the gas

State variables : Volume, Temperature

Process variables : Internal energy, work done by the gas

State variables : work done by the gas, heat rejected by the gas.

Process variables : Temperature, volume

State variables : Internal energy, volume

Process variables : Work done by the gas, heat absorbed by the gas

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Explanation

Internal energy and volume depend upon states.

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One mole of a real gas is subjected to heating at constant volume from $(P_{1}, V_{1}, T_{1})$ state to $(P_{2}, V_{2}, T_{2})$ state. Then it is subjected to irreversible adiabatic compression against constant external pressure of P₃ atm till syatem reaches the final state $(P_{3}, V_{3}, T_{3})$ . If the constant volume molar heat capacity of real gas is C_V. Find out correct expression for $∆ H$ from state 1 to state 3-

$C_{v} (T_{3} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})$

$C_{v} (T_{2} - T_{1}) + (P_{3} V_{2} - P_{1} V_{1})$

$C_{v} (T_{2} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})$

$C_{p} (T_{2} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})$

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Explanation

$∆ H_{I} = C_{v} (T_{2} - T_{1}) + (P_{2} V_{1} - P_{1} V_{1})$

$∆ H_{I I} = ∆ U + P_{3} V_{2} - P_{2} V_{1} = P_{3} (V_{2} - V_{1}) + P_{3} V_{2} - P_{2} V_{1} = P_{3} V_{1} - P_{2} V_{1} ∆ H_{o v e r a l l} = C_{v} (T_{2} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})$

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$[\begin{matrix} H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (g); ∆ H = 241.8 k J \\ C O (g) + \frac{1}{2} O_{2} (g) = C O_{2} (g); ∆ H = 283 k J \end{matrix}]$

The heat evolved in the combustion of 112 litres of water gas(mix of equal volumes of H₂ and CO)

241.8 kJ

283 kJ

1312 kJ

1586 kJ

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Explanation

112 litre (at NTP) contains no. of moles = $\frac{112}{22.4}$ =5

$∴$ 2.5 moles of H₂ and 2.5 moles of CO present. Hence total amount of heat evolved

$= 2.5 \times 241.8 + 2.5 \times (283) = 1312 k J$

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In which case of mixing of a strong acid and a base, each of 1(N) concentration, temperature-increase is the highest ?

20 ml acid and 30 ml alkali

10 ml acid and 40 ml alkali

25 ml acid and 25 ml alkali

35 ml acid and 15 ml alkali

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Explanation

When 25 ml acid is reacting with 25 ml of alkali, maximum equivalent of acid is reacting with alkali since the volume of the resulting solution is the same in every cases, therefore, increase in temperature is the highest where maximum equivalents of acids and alkalies are reacting.

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The heats of neutralisation of four acids A, B, C and D are -13.7, -9.4, -11.2 and -12.4 kcal respectively when they are neutralised by a common base. The acidic character obeys the order :

A>B>C>D

A>D>C>B

D>C>B>A

D>B>C>A

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Explanation

Lower is heat of neutralisation, more is dissociation energy, weaker is acid

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The $∆ H_{f}^{\circ}$ for CO₂(g), CO(g) and H₂O(g) are -393.5, -110.5 and -241.8 kJ mol^-1 respectively. The standard enthalpy change (in kJ) for the reaction,

$C O_{2} (g) + H_{2} (g) \to H_{2} O (g)$ is :

524.21

41.2

-262.5

-41.2

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Explanation

Given

$C + O_{2} \to C O_{2}; ∆ H^{\circ} = - 393.5 k J . . . . . . . (i) C + (1 / 2) O_{2} \to C O; ∆ H^{\circ} = - 110.5 k J . . . . . . . (i i) H_{2} + (1 / 2) O_{2} \to H_{2} O; ∆ H^{\circ} = - 241.8 k J . . . . . . . (i i i)$

By (ii) + (iii) - (i),

$C O_{2} + H_{2} \to C O + H_{2} O; ∆ H^{\circ} = + 41.2$

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