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

C2H4(g) + 6F2(g) → 2CF4(g) + 4HF(g)

H2(g) + F2(g) → 2HF(g)       ∆H1°= -537 kJ

C(s) + 2F2(g) →CF4(g)         ∆H2°=-680 kJ

2C(s) + 2H2(g) →C2H4(g)    ∆H3°= 52 kJ

1. –1165 kJ

2. –2486 kJ

3. +1165 kJ

4. +2486 kJ

Subtopic:  Enthalpy & Internal energy |
 83%
Level 1: 80%+
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Entropy decreases during:

1. Crystallization of sucrose from solution

2. Rusting of iron

3. Melting of ice

4. Vaporization of camphor

Subtopic:  Spontaneity & Entropy |
 85%
Level 1: 80%+
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The volume versus temperature graph for two moles of monoatomic gas is shown in the figure. The ratio of work done by the gas to the heat absorbed by it in the process \(A\) to \(B\) is:

 
1. \(\dfrac{1}{2}\) 2. \(\dfrac{2}{5}\)
3. \(\dfrac{3}{7}\) 4. \(\dfrac{3}{5}\)
Subtopic:  Work Done by a Gas |
 79%
Level 2: 60%+
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The \((P\text-V)\) graph of an ideal monoatomic gas is as shown. The molar heat capacity of gas will be:

          
1. \(2R\)
2. \(3R\)
3. \(5R\)
4. \(7R\)

Subtopic:  Molar Specific Heat |
 54%
Level 3: 35%-60%
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A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0ºC. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be-
(R = 8.314 J/mol K) (ln 7.5 = 2.01)

1. q = +208 J, w = -208 J 2. q = -208 J, w = -208 J
3. q = -208 J, w = + 208 J 4. q = +208 J, w = + 208 J
Subtopic:  First Law of Thermodynamics |
 83%
Level 1: 80%+
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∆Hf0 (298K) of methanol is given by the chemical equation:

1. \(\mathrm{C}(\text { diamond })+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_3 \mathrm{OH}_{(\mathrm{l})}\)

2. \(\mathrm{CH}_{4(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_3 \mathrm{OH}_{(\mathrm{g})}\)

3. \(\mathrm{CO}_{(\mathrm{g})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_3 \mathrm{OH}_{(\mathrm{l})}\)

4. \(\mathrm{C}(\text { graphite })+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_3 \mathrm{OH}_{(\mathrm{l})}\)

Subtopic:  Thermochemistry |
 78%
Level 2: 60%+
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When 4 g of iron is burnt in oxygen to form ferric oxide at constant pressure, 29.28 kJ of heat is evolved. Calculate the enthalpy of formation of ferric oxide.

(At. mass of Fe = 56) ?

1. -81.98 kJ

2. - 819.8 kJ

3. - 40.99 kJ

4.  +819.8 kJ

Subtopic:  Thermochemistry |
 59%
Level 3: 35%-60%
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Reversible expansion of an ideal gas under isothermal and adiabatic conditions are shown in the figure:

AB→Isothermal expansion

AC→Adiabatic expansion

Which of the following options is not correct?

1. \(\Delta S_{\text {isothermal }}>\Delta S_{\text {adiabatic }} \) 2. \(T_A=T_B \)
3. \(W_{\text {isothermal }}>W_{\text {adiabatic }} \) 4. \(T_C>T_A\)
Subtopic:  2nd & 3rd Law of Thermodynamics |
 55%
Level 3: 35%-60%
NEET - 2019
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An ideal gas expands isothermally at 300 K from a volume of 10⁻³ m³ to 10⁻² m³ against a constant
external pressure of 10⁵ N m⁻². Calculate the work associated with this expansion using the chemistry sign convention.

1. \(+270 ~kJ\) 2. \(–900 ~J\)
3. \(+900 ~kJ\) 4. \(–900~ kJ\)
Subtopic:  First Law of Thermodynamics |
 72%
Level 2: 60%+
NEET - 2019
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Hydrolysis of sucrose is given by the following reaction

Sucrose + H2O ⇌ Glucose + Fructose

If the equilibrium constant (Kc) is 2×1013 at 300 K, the value of ∆rG⊝ at the same temperature will be:

1. 8.314 J mol–1 K–1×300 K×ln (2×1013)

2. 8.314 J mol–1 K–1×300 K×ln (3×1013)

3. –8.314 J mol–1 K–1×300 K×ln (4×1013)

4. –8.314 J mol–1 K–1×300 K×ln (2×1013)

Subtopic:  Gibbs Energy Change |
 81%
Level 1: 80%+
NEET - 2020
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