Vapour density of the equilibrium mixture of the reaction SO₂Cl₂(g)\(\rightleftharpoons\) SO₂(g) + Cl₂(g) is 50.0. Percent dissociation of SO₂Cl₂ is :

33.33
35.0
30.0
66.67

Solution

SO₂Cl₂(g)\(\rightleftharpoons\) SO₂(g) + Cl₂(g)

α%=\(\frac{D-d}{d(y-1)}\times 100\) (y= 2)

\(D=\frac{molar\: mass\: of\: SO_{2}Cl_{2}}{2}=\frac{135}{2}=67.5;d=50.0\)
(given)

\(∴\alpha =\frac{67.5-50.0}{50.0(2-1)}\times 100=35\)%

A solution having hydrogen ion concentration is 0.0005 geqvt./litre, its pOH is :

8.2798
10.6990
12.7854
13.3344

Solution

pH =– log [H⁺] = – log [5 × 10^-4]
= 4 –log 5 =4 – [log 10– log 2]
= 3 +log 2 = 3.3010
pOH = 14– pH = 14 – 3.3 =10.7.

If α is the fraction of HI dissociated at equilibrium in the reaction, 2 HI (g)\(\rightleftharpoons\) H₂ (g) + I₂ (g), starting with 2 moles of HI, the total number of moles of reactants and products at equili brium are

2 + 2α
2
1 +α
2– α

Solution

According to equation

Total moles at equilibrium
= 2 – 2α + α + α = 2 mole

For homogeneous gaseous reaction,
4NH₃+ 5O₂→4NO + 6H₂O
The units of equilibrium constant K_c will be –

mol lit^-1
mol^-1lit^-1
mol²lit^-2
K_c is unitless

Solution

Unit of K_C ⇒(mol lit^-1)^Δn
For given reaction,
Δn = (4 + 6) – (4 + 5) = 1,
K_C→mol lit^-1

5 mole of NH₄HS (s) start to decompose at a particular temperature in a closed vessel. If pressure of NH₃(g) in the vessel is 2 atm, then Kp for the reaction,
NH₄HS (s)\(\rightleftharpoons\)NH₃(g) + H₂S (g),will be –

Solution

P_NH3=P_H2S=2 atm
∴ K_P=P_PH3×P_H2S=4

One mole of CH₃COOH and one mole of C₂H₅OH reacts to produce ²⁄₃ mol of CH₃COOC₂H₅. The equilibrium constant is :

2
+ 2
– 4
+ 4

Solution

CH₃COOH + C₂H₅OH→CH₃COOC₂H₅ + H₂O

At equb. 1-²⁄₃ 1-²⁄₃ ²⁄₃ ²⁄₃

Let the total volume = V L

∴[CH₃OOH]=¹⁄₃ V mol/L

[C₂H₅OH ] =¹⁄₃ V mol/L

[CH₃COOC₂H₅] =²⁄₃ V mol/L

[H₂O] =²⁄₃ V mol/L

\(∴K=\frac{\frac{2}{3}V\times \frac{2}{3}V}{\frac{1}{3}V\times \frac{1}{3}V}=4\)

PCl₅ is dissociating 50% at 250°C at a total pressure of Patm. If equilibrium constant is Kp, then which of the following relation is numerically correct –

Kp= 3P
P= 3Kp
P= 23PK
Kp= 23

Solution

PCl₅\(\rightleftharpoons\)PCl₃+Cl₂
Moles at equilibrium
¹⁄₂ ¹⁄₂ ¹⁄₂
Mole fraction at equilibrium
¹⁄₃ ¹⁄₃ ¹⁄₃
Partial pressure at equilibrium
^P⁄₃ ^P⁄₃ ^P⁄₃

\(K_{P}=\frac{\frac{P}{3}\times \frac{P}{3}}{P/3}=\frac{P}{3}\)

For a chemical reaction;
A (g) +B (l)\(\rightleftharpoons\)D (g) + E (g)
Hypothetically at what temperature, Kp = Kc(when, R = 0.08 l-atm/mole-K)

T =0 K
T = 1K
T = 12.5 K
T = 273 K

Solution

As Kp= Kc RT\(^{\Delta ng}\)
Here Δng = 1
So, Kp= Kc
when RT = 1 Thus T = 12.5 K

The value of K_P for the reaction:
2H₂S(g)=2H₂(g)+S₂(g) is 1.2×10^-2 at 1065°C. The value for K_C is:

<1.2×10^-2
>1.2×10^-2
1.2×10^-2
0.12×10^-2

Solution

For the given reaction,

2H₂S(g) ⇌ 2H₂(g)+S₂(g)

Δn=3-2=1

K_c=\(\frac{K_{P}}{(RT)^{n}}=\frac{1.2\times 10^{-2}}{(RT)^{1}}\)

∴ K_C<K_P or K_C < 1.2×10^-2

Pure ammonia is placed in a vessel at a temperature where its dissociation constant (a) is appreciable. At equilibrium :

K_P does not change significantly with pressure.
does not change with pressure.
concentration of NH₃ does not change with pressure.
concentration of hydrogen is less than that of nitrogen.

Solution

Statement (a) is correct and the rest statements are wrong. K_P depends only on temperature hence at constant temp. K_P will not change.

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