Chemistry

11.45 Anhydride of carbonic acid is:

(a) H₂CO₃ (b) CO₃ (c) CO^- (d) CO₂

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(d) CO₂ is regarded as anhydride of carbonic acid.

H₂CO₃ ————> H₂O + CO₂

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MCQs

11.44 Which of the following statement is incorrect?

(a) Non-metals and metalloids exist only in the p-block of the periodic table.

(b) The non-metallic character of elements decreases down the group.

(c) The change of non-metallic to metallic character can be best illustrated by the nature of oxides they form.

(d) The non-metal oxides are basic whereas metal oxides are acidic or neutral in nature

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(d) is an incorrect statement because non-metal oxides are acidic or neutral whereas metal oxides are basic in nature.

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11.43. Assertion: Boron halides form addition compounds with NH₃.

Reason: Boron halides are Lewis bases.

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(c) Boron halides are Lewis acids and can accept a pair of electrons from amines to form addition product.

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11.42. Assertion: Heavier elements form pπ– pπ bonds.

Reason: Their atomic orbitals are too large and diffuse to have effective overlapping.

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(d) Heavier elementsdo not form pπ– pπ bonds because their atomicorbitals are too large and diffuse to haveeffective overlapping.

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Assertion and Reason:

Directions:

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(c) Assertion is correct statement but reason is wrong statement.

(d) Assertion is wrong statement but reason is correct statement.

11.39. Assertion: Non-metals readily form anions, in contrast to metals whichreadily form cations.

Reason: In general, non-metals have higher ionisation enthalpies and higherelectronegativities than the metals.

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4.39 The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

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4.39 Given, k₂ = 4k₁, T₁ = 293K and T₂ = 313K

We know that from the Arrhenius equation, we obtain

On solving, we get,

E_a = 58263.33 J mol^-1 or 58.26 kJ mol^-1

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4.38 The time required for 10% completion of a first order reaction at 298K is equal to that required for its 25% completion at 308K. If the value of A is 4 * 1010s ^–1, calculate k at 318K and Ea.

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4.38 We know, time t = (2.303/k) * log ( [R]₀/ [R])

Where, k- rate constant

[R]₀-Initial concentration

[R]-Concentration at time 't'

At 298K, If 10% is completed, then 90% is remaining. t = (2.303/k) * log ( [R]₀/0.9 [R]₀)

t = (2.303/k) * log (1/0.9) t = 0.1054 / k

At temperature 308K, 25% is completed, 75% is remaining t' = (2.303/k') * log ( [R]₀/0.75 [R]₀)

t' = (2.303/k') * log (1/0.75) t' = 2.2877 / k'

But, t = t'

0.1054 / k = 2.2877 / k' / k = 2.7296

From Arrhenius equation, we obtain log k₂/k₁ = (E_a / 2.303 R) * (T₂ - T₁) / T₁T₂

Substituting the values,

E_a = 76640.09 J mol^-1 or 76.64 kJ mol^-1 We know, log k = log A –E_a/RT

Log k =

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4.37 The decomposition of A into product has value of k as 4.5* 103 s^–1 at 10°C and energy of activation 60 kJ mol^–1. At what temperature would k be 1.5 * 104s^–1?

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4.37 From Arrhenius equation, we obtain

Also, k₁ = 4.5 * 10³ s ^-¹

T₁ = 273 + 10 = 283 K

k₂ = 1.5 * 10⁴ s ^-¹

E_a = 60 kJ mol ^-¹ = 6.0 * 10⁴ J mol ^-¹

Then,

→ 0.5229 = 3133.627 * (T₂-283)/ (283 * T₂)

→ 0.0472T₂ = T₂-283 T₂ = 297K or T₂ = 24⁰ C

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4.36 The rate constant for the first order decomposition of H₂O₂ is given by the following equation: log k = 14.34 – 1.25 * 104K/T. Calculate E_a for this reaction and at what temperature will its half-period be 256 minutes?

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4.36 We know, The Arrhenius equation is given by k = Ae^-Ea/RT Taking natural log on both sides,

Ln k = ln A- (E_a/RT)

Thus, log k = log A - (E_a/2.303RT). eqn 1

The given equation is log k = 14.34 – 1.25 * 10⁴K/T. eqn 2

Comparing 2 equations, E_a/2.303R = 1.25 * 10⁴K

E_a = 1.25 * 10⁴K * 2.303 * 8.314

E_a = 239339.3 J mol^-1 (approximately) E_a = 239.34 kJ mol^-1

Also, when t_1/2 = 256 minutes,

k = 0.693 / t_1/2

= 0.693 / 256

= 2.707 * 10^-3 min^-1 k = 4.51 * 10^-5s^–1

Substitute k = 4.51 * 10^-5s^–1 in eqn 2,

log 4.51 * 10^-5 s^–1 = 14.34 – 1.25 * 10⁴K/T

log (0.654-5) = 14.34– 1.25 * 10⁴K/T = 1.25 * 10⁴/ [ 14.34- log (0.654-5)] T = 668.9K or T =

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4.35 The decomposition of hydrocarbon follows the equation k = (4.5 * 10¹¹s^–1) e^-28000K/T. Calculate E_a

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