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All Physics Chemistry Botany Zoology

Language English हिंदी

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Identify the correct statement for change of Gibbs energy for a system ( $∆$ G_system) at constant temperature and pressure:

If $∆$ G_system > 0, the process is spontaneous

If $∆$ G_system = 0, the system has attained equilibrium

If $∆$ G_system = 0, the system is still moving in a particular direction

If $∆$ G_system < 0, the process is not spontaneous

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Explanation

If the Gibbs free energy for a system ( $∆$ G_system) is equal to zero, then the system is present in equilibrium at a constant temperature and pressure.

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The enthalpy and entropy change for the reaction :

Br₂(l)+Cl₂(g) $\to$ 2BrCl(g) are 30 kJ mol^-1 and 105 JK^-1 mol^-1 respectively.

The temperature at which the reaction will be in equilibrium is :

285.7 K

273 K

450 K

300 K

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Explanation

At equilibrium Gibbs free energy change (ΔG°) is equal to zero. The following thermodynamic relation is used to show the relation of ΔG° with enthalpy change (ΔH°) and entropy change(ΔS°)

ΔG° = ΔH°-TΔS

0 = 30 x 10³ (J mol^-1) - T x 105 (J K^-1 mol^-1)

T = 3X10³/105 K= 285.71 K

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Consider the reactionat 300K

H₂₍₉₎ + Cl₂₍₉₎ →2HCI_(g), ΔH° = — 185 KJ

If 3 mole of H completely react with 3 mol of Cl₂ to form Cl, $∆$ U° of the reaction will be

Zero

–185 KJ

-555 KJ

None

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Explanation

Δng=0, ΔH° = ΔU° = – 185 KJ

For 3 mole, ΔU° = 3 x (– 185) = – 555 KJ

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Fora perfectly crystalline solid C_pm = aT³, where a is constant. If C_pm is 0.42 J/K–mol at 10 K, molar entropy at 10 K is

0.42 J/K–mol

0.14 J/K–mol

4.2 J/K–mol

zero

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Explanation

0.42 = a(10)³⇒a = 0.42 × 10³

S_m = $\int_{0}^{10} \frac{C_{pm}}{T}$ dT = $\int_{0}^{10} {aT}^{2} = \frac{a}{3} [10^{3} - 0] = \frac{0 .42}{3} = 0 .14 J / K - mol$

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One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :

[R = 0.082 L atm mol^–1 K^–1 = 8.3 J mo1^–1K^–1].

Rln (24.6)

Rln (2490)

$\frac{3}{2}$ Rln(24.6)

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Explanation

ΔS = nR ln $(\frac{V_{f}}{V_{i}}) = R \ln (\frac{P_{i}}{P_{f}})$ = R ln $(\frac{300 R}{1 L \times 1 atm})$ = Rln (24.6)

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At 1000 K water vapour at 1 atm. has been found to be dissociated into H₂ and O₂ to the extent of 3 x 10^–6 %.Calculate the free energy decrease of the system, assuming ideal behaviour.

–ΔG = 90,060 cal

–ΔG = 20 cal

–ΔG = 480 cal

–ΔG = –45760 cal

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Explanation

H₂O(g) $\overset{}{⇌}$ H₂O + $\frac{1}{2}$ O₂

1 – α α α/2

K_P = $\frac{α \times {(α / 2)}^{1 / 2}}{(1 - α)}$ × $\frac{P}{(1 + \frac{α}{2})}$ = 11.62 × 10–11 (atm)^1/2

ΔG° = – RT In K_p = – 45.76 kcal

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When 1 mole of an ideal gas at 20 atm pressure and 15 L volume expands such that the final pressure becomes 10 atm and the final volume become 60 L. Calculate entropy change for the process (C_pm = 30.96 J mole^–1 K^–1)

80.2 J.k^–lmo1^–1

62.42 kJ.k^–1 mol^–1

120 x 10² J1c¹mol^–1

27.22 J.k^–1mor

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Explanation

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

$\frac{T_{2}}{T_{1}} = \frac{6}{3}$

ΔS = 2.303 × n $\{C_{p} \log_{10} \frac{T_{2}}{T_{1}} + R \log_{10} \frac{P_{1}}{P_{2}}\}$

ΔS = 2.303 × 1 $\{10 .96 \log_{10} \frac{6}{3} + R \log_{10} \frac{20}{10}\}$

ΔS = 27.22 J.K^–1 mol^–1

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During winters, moisture condenses in the form of dew and can be seen on plant leaves and grass. The entropy of the system in such cases decreases as liquids possess lesser disorder as compared to gases. With reference to the second law, which statement is correct, for the above process ?

The randomness of the universe decreases

The randomness of the surroundings decreases

Increase is randomness of surroundings equals the decrease in randomness of system

The increase in randomness of the surroundings is greater as compared to the decrease in randomnessof the system.

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Explanation

As dew formation is spontaneous process therefore entropy or randomness of the universe will increase. As randomeness of system has decreased but randomness of the surrounding will increase larger so that change is positive.

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For which process will $∆$ H° and $∆$ G° be expected to be most similar.

2Al(s) + Al₂O₃(s) → 2Fe(s) + A_l2O₃(s)

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

2NO₂(g) → N₂O₄(g)

2H₂(g) + O₂(g) → 2H₂O(g)

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Explanation

All product and reactants are in solid state

So ΔS° ≈ 0

ΔG° = ΔH – TΔS°

So ΔG° ≈ ΔH°

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Heat produced in calories by the combustion of one gram of carbon is called

Heat of combustion of carbon

Heat of formation of carbon

Calorific value of carbon

Heat of production of carbon

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Explanation

It is the definition of calorific value.

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Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.