Determine which of the following reactions at constant pressure represent surrounding that do work on the system?
I. $4 N H_{3} (g) + 70_{2} (g) \to 4 N O_{2} (g) + 6 H_{2} O (g)$
II. $C O (g) + 2 H_{2} (g) \to C H_{3} O H (l)$
III. $C (s, g r a p h i t e) + H_{2} O (g) \to C O (g) + H_{2} (g)$
IV. $H_{2} O (s) \to H_{2} O (l)$

III, IV

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II and III

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II, IV

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I and II, IV

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Solution

The correct option is C I and II, IV
When a gas is evolved during a chemical reaction, the gas can be imagined as displacing the atmosphere - pushing it back against the atmospheric pressure. The work done is therefore $V \times P$ where V is the volume of gas evolved, and P is the atmospheric pressure.

$4 N H_{3} (g) + 7 O_{2} (g) \to 4 N O_{2} (g) + 6 H_{2} O (g)$
In this reaction 11 vols of reagent combine to form 10 vol of product. The total volume falls, so the work done is negative (the atmosphere does work on the gaseous system).

$C O (g) + 2 H_{2} (g) \to C H_{3} O H (l)$
3 volumes of reagent combine to form 1 vol of product. The total volume falls, so the work done is negative (the atmosphere does work on the gaseous system).

$C (s, g r a p h i t e) + H_{2} O (g) \to C O (g) + H_{2} (g)$
Here, 2 vols of reagents combine to produce 2 vols of product. No change in surrounding.

$H_{2} O (s) \to H_{2} O (l)$
In this reaction solid volumes of reagent combine to form liquid product. The total volume increases, so the work done on the atmosphere is negative.

Suggest Corrections

Similar questions

Q. Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i)

{C O C l}_{2} (g) ⇌ C O (g) + {C l}_{2} (g)

(ii)

{C H}_{4} (g) + 2 S_{2} (g) ⇌ {C S}_{2} (g) + 2 H_{2} S (g)

(iii)

{C O}_{2} (g) + C (s) ⇌ 2 C O (g)

(iv)

2 H_{2} (g) + C O (g) ⇌ {C H}_{3} O H (g)

(v)

{C a C O}_{3} (s) ⇌ C a O (s) + {C O}_{2} (g)

(vi)

4 {N H}_{3} (g) + 5 O_{2} (g) ⇌ 4 N O (g) + 6 H_{2} O (g)

Q. In which of the following changes at constant pressure is work done by system on surrounding?

	Initial state		Final state
(i)	$H_{2} O (g)$	$\to$	$H_{2} O (l)$
(ii)	$H_{2} O (s)$	$\to$	$H_{2} O (g)$
(iii)	$H_{2} O (l)$	$\to$	$H_{2} O (s)$
(iv)	$3 H_{2} (g) + N_{2} (g)$	$\to$	$2 N H_{3} (g)$
(v)	$C a C O_{3} (s)$	$\to$	$C a O + C O_{2} (g)$

Calculate the equilibrium constant ( $K_{p}$ ) for the reaction, $C (s) + C O_{2} (g) ⇌ 2 C O (g)$ , at $1300 K$ from the following data:
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); K_{p} (1300 K) = 3.9 a t m$

$H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g); K_{p} (1300 K) = 0.623 a t m$

Q. Determine which of the following reactions taking place at constant pressure represent system that do work on the surrounding environment

(i)

{A g}^{+} (a q) + {C I}^{-} (a q) ⟶ A g C I (s)

(ii)

{N H}_{4} C I (s) ⟶ {N H}_{3} (g) + H C I (g)

(iii)

2 {N H}_{3} (g) ⟶ N_{2} (g) + 3 H_{2} (g)

Q. For the reaction,

C (s) + H_{2} O (g) ⇌ H_{2} (g) + C O (g)

, by increasing pressure following equilibrium :

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Determine which of the following reactions at constant pressure represent surrounding that do work on the system?I. 4NH3(g)+702(g)→4NO2(g)+6H2O(g)II. CO(g)+2H2(g)→CH3OH(l)III. C(s,graphite)+H2O(g)→CO(g)+H2(g)IV. H2O(s)→H2O(l)