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Assume each reaction is carried out in an open container. For which reaction will $∆$ H = $∆$ E ?

H2(g) + Br₂(g) $\to$ 2HBr(g)

C(s)+ 2H₂O(g) $\to$ 2H₂(g) + CO₂(g)

PCl₅(g) $\to$ PCl₃(g) + Cl₂(g)

2CO(g)+ O2(g) $\to$ 2CO₂(g)

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Explanation

As we know that

$∆$ H = $∆$ E + P $∆$ V

or $∆$ H = $∆$ E + $∆$ nRT

where $∆$ H $\to$ change in enthalpy of system (standard heat at constant pressure)

$∆$ E $\to$ Change in internal energy of system (Standard heat at constant volume)

An $\to$ no. of gaseous moles of product - no. of gaseous moles of reactant

R $\to$ gas constant

T $\to$ absolute temperature

If $∆$ n = 0 for reactions which is carried out in an open container, therefore $∆$ H = $∆$ E

So for reaction (1) $∆$ n = 2 - 2 = 0

Hence, for reaction (1), $∆$ H = $∆$ E

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For the reaction,

CH₄(g) + 2O₂(g) $⇌$ CO₂(g) + 2H₂O (l),

$∆$ _rH = -170.kJ mol-1

Which of the following statements is not true?

At equilibrium, the concentrations of CO₂(g) and H₂O(l) are not equal

The equilibrium constant for the reaction is given by K_p = [CO₂]/CH₄][O₂]

Addition of CH₄(g) or O₂ (g) at equilibrium will cause a shift to the right

The reaction is exothermic

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Explanation

For the reaction,

CH₄ (g) + 2O₂(g) $⇌$ CO₂ (g) + 2H₂O (l) $∆$ H_r = -170.8 kJ mol^-1

This equilibrium is an example of heterogeneous chemical equilibrium. Hence, for it

K_c = [CO₂]/[CH₄][O₂]² ....(i)

(Equilibrium constant on the basis of conc.)

Thus, in it concentration of CO₂ (g) and H₂O (l) are not equal at equilibrium. The equilibrium constant (K_p) = [CO₂]/[CH₄][O₂] is not correct expression. In addition of CH₄(g) or O₂ (g) at equilibrium, K_c will be decreased according to expression (i) but Kc remains constant at constant temperature for a reaction, so for maintaining the constant value of K_c, the concentration of CO₂ will increase in same order. Hence on an addition of CH₄ or O₂, equilibrium will cause to the right. This reaction is an example of an exothermic reaction.

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Which of the following pairs constitutes a buffer?

HNO₂ and NaNO₂

NaOH and NaCl

HNO₃ and NH4NO₃

HCl and KCl

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Explanation

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A pair constituent with a HNO₂ and NaNO₂ because HNO₂ is weak acid and NaNO₂ is a salt of the weak acid (HNO₂) with strong base (NaOH). Hence, it is an example of acidic buffer solution.

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The hydrogen ion concentration of a 10^-8 M HCl aqueous solution at 298 K(K_w 10^-14)is:

1.0×10^-6 M

1.0525 x 10^-7 M

9.525 x 10^-8 M

1.0×10^-8 M

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Explanation

(b) In aqueous solution of 10^-8 M HCl, [H⁺] is based upon the concentration of H⁺ ion of water 10^-8 M HCl and concentration of H⁺ ion of water K_w of H₂O=10^-14=[H⁺][OH^-] or[H⁺]=10^-7 M (due to its neutral behaviour)

So, in aqueous solution of 10^-8 M HCI,

$[H^{+}] = [H^{+}] of HCl + [H^{+}] of water = 10^{- 8} + 10^{- 7} = 11 \times 10^{- 8} M = 1.10 \times 10^{- 7} M$

Hence, answer is nearer to (b).

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Two flasks A and B of equal volume containing 1 mole and 2 mole of O₃ respectively, are heated to the sametemperature. When the reaction 2O₃ $⇋_{}^{}$ 3O₂ practically stops, then both the flasks shall have

the same ratio [O₂] /[O₃]

the same ratio : [O₂]³/²/[O₃]

only O₂

the same time to reach equilibrium

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Explanation

For reaction

$\frac{[O_{2}]^{3}}{[O_{3}]^{2}} =$ constantso $\sqrt{K_{C}}$ = $\frac{[O_{2}]^{3 /2}}{[O_{3}]}$ = constant

= will be same for both the containers

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A 10L container at 300K contains CO₂ gas at pressure of 0.2 atm and an excess solid CaO (neglect thevolume of solid CaO). The volume of container is now decreased by moving the movable piston fitted in the container. What will be the maximum volume of container when pressure of CO₂ attains its maximum value given that CaCO₃ (s) $⇋_{}^{}$ CaO(s) + CO₂(g)K_ph = 0.800 atm

5 L

2.5 L

1 L

The information is insufficient.

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Explanation

K_p = 0.800 atm = $P_{{CO}_{2}}$ = maximum pressure of CO₂ in the container to calculate maximum volumeof container the $P_{{CO}_{2}}$ = 0.8 atm and none of $P_{{CO}_{2}}$ = 0.8 should get converted into CaCO₃(s).

so V(0.800 atm) = (10 L) (0.2 atm)

so v = 2.5 L

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At a certain temperature the following equilibrium is established, CO(g) + N0₂(g) $\overset{}{⇌}$ CO₂(g)+ NO(g)One mole of each of the four gas is mixed in one litre container and the reaction is allowed to reach equilibrium state. When excess of baryta water (Ba(OH)₂) is added to the equilibrium mixture, the weight of white ppt (BaCO₃) obtained is 236.4 gm. The equilibrium constant K_c of the reaction is (Ba = 137)

1.2

2.25

2.1

3.6

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Explanation

	CO(g)	+	NO₂(g)	$\overset{}{⇌}$	CO₂(g)	+	NO(g)
t =	0.1 mole		1 mole		1 mole		1 mole
Al eq.	1 – x		1 – x		1 + x		1 + x

CO₂ + Ba(OH)₂ $\overset{}{\to}$ BaCO₃

mole of BaCO₃ = $\frac{236 .4}{197}$ = 1.2

So mole of CO₂ at eq. = 1.2

or 1 + × = 1.2

x = 0.2

K_C = ${(\frac{1 + x}{1 - x})}^{2} = {(\frac{1 .2}{0 .8})}^{2} =$ 2.25

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At temperature T^, the compound AB₂ (g) dissociates according to the reaction, 2AB₂ (g) $\overset{}{⇌}$ 2 AB(g) + B₂(g). With a degree of dissociation x^, which is small compared with unity. Deduce the expression for x in terms of the equilibrium constant, K_p and the total pressure, P.

$x = {[\frac{2 K_{p}}{P}]}^{- 1 /3}$

$x = {[\frac{K_{p}}{P}]}^{1 /3}$

$x = {[\frac{K_{p}}{2P}]}^{1 /3}$

$x = {[\frac{6 K_{p}}{3 P}]}^{1 /3}$

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Explanation

2AB₂ (g) $\to_{}^{}$ 2AB (g) + B₂ (g)

Mole before dissociation 1 0 0

Mole after dissociation (1 – x) x $\frac{x}{2}$

Total mole at equilibrium (Σn) = 1 - x + x + $\frac{x}{2}$ = 1 + $\frac{x}{2}$

Now, K_p = $\frac{n_{B_{2}} \times {(n_{AB})}^{2}}{{(n_{{AB}_{2}})}^{2}} \times {[\frac{P}{\sum n}]}^{Δn}$

K_p = $\frac{\frac{x}{2} . (x)^{2}}{{(1 - x)}^{2}} \times {[\frac{p}{1 + \frac{x}{2}}]}^{1}$

K_p = $\frac{x^{3} p}{2}$ $[∵ x is small, ∴ 1 - x and 1 + \frac{x}{2} = 1]$

x = $^{3} \sqrt{\frac{2 K_{p}}{p}}$

$x = {[\frac{K_{p}}{2P}]}^{1 /3}$

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Equilibrium constant for the given reaction is kc = 10²⁰ at temperature 300 K,A(s) + 2B (aq.) $\overset{}{⇌}$ 2C (s) + D (aq.)K = 10²⁰ The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mole/litre of B at 300 K is

$~ 4 \times 10^{- 11}$

$~ 2 \times 10^{- 10}$

$~ 2 \times 10^{- 11}$

$~ 10^{- 10}$

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Explanation

A(s) + 2B(aq) $\to_{}^{}$ 2C(s) + D(aq)

Initial 1 $\frac{1}{3}$ 0 0

At eq. 1 – x $\frac{1}{3} - 2 x$ 2x x

$\approx$ a

x $\approx$ 1/3

10²⁰ = $\frac{\frac{1}{3}}{{[B]}^{2}}$

10²⁰ = $\frac{\frac{1}{3}}{a^{2}}$

a² = $\frac{1}{3 \times 10^{20}}$ = $\frac{10^{- 20}}{3}$

a = $\frac{10^{- 10}}{\sqrt{3}} \approx$ 4 × 10^–11 M

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10l box contain O₃ and O₂ at equilibrium at 2000 K. The ΔG° = –534.52 kJ at 8 atm equilibrium pressure.The following equilibrium is present in the container. 2O₃(g) $\overset{}{⇌}$ 3O₂(g). The partial pressure of O₃ will be (In 10 = 2.3, R = 8.3 Jmole^–1K^–1):

8 ×10⁴

22.62 × 10^-7

9.71 × 10^-6

9.71 × 10^-2

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Explanation

ΔG° = – RT In K = – 2.3 × 2000 × 8.3 log K

$\frac{534 .52 × 10^{3}}{2 .3 × 8 .3 × 2000} = \log K$

K = 10¹⁴

2O₃(g) $\overset{}{⇌}$ 3O₂(g) K = 10¹⁴

So $P_{O_{2}}$ <<< $P_{O_{2}}$

So $P_{O_{2}}$ + $P_{O_{2}}$ = 8

$P_{O_{2}}$ = 8 atm

K = 10¹⁴ = $\frac{{(P_{O_{2}})}^{3}}{{(P_{O_{3}})}^{2}} = \frac{{(8)}^{3}}{P_{O_{3}}}$

$P_{O_{3}}$ = 22.62 × 10^–7 atm.

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