You visited us 57 times! Enjoying our articles? Unlock Full Access!

Dalton's Law of Partial Pressures

In a 13L ve...

Question

In a $13 L$ vessel initially following reaction occur $C (s) + S_{2} (g) ⇌ {C S}_{2} (g)$ by $12 g$ $C$ , $64 g$ $S_{2}$ , $76 g$ ${C S}_{2}$ at $1027^{o} C$ temperature then total pressure is:

200 R

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

158 R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

79 R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $200 R$
$C (s) + S_{2} (g) ⇌ C S_{2} (g)$
Number of moles of $C = \frac{12 g}{12 g} = 1$
Number of moles of $S O_{2} = \frac{64}{64} = 1$
Number of moles of $C S_{2} = \frac{76}{76} = 1$
Total number of gaseous moles= $1 (S_{2}) + 1 (C S_{2}) = 2$ moles
From ideal gas equation, $P V = n R T$
$P = \frac{n R T}{V}$
$= \frac{2 \times R \times (273 + 1027)}{13}$
$\Rightarrow \frac{2 R (1300)}{13}$
$= 200 R$

Suggest Corrections

Similar questions

13 L

C (s) + S_{2} (g) ⇌ {C S}_{2} (g)

12 g

C

64 g

S_{2}

76 g

{C S}_{2}

1027^{o} C

C (s) + S_{2} (g) ⇌ C S_{2} (g)

12 g = C,

64 g = S_{2}

76 g = C S_{2}

1027^{o} C

13 L

C (s) + S_{2} (g) ⇌ C S_{2} (g)

12 g C, 64 g S_{2},

76 g C S_{2}

1027^{\circ} C

13

C (s) + S_{2} (g) \to C S_{2}

12

S_{2}

76

C S_{2}

1027

9.40

900^{o} C

S_{2} (g) + C (s) ⇌ C S_{2} (g)

1.42 a t m

S_{2}

C (s)

Join BYJU'S Learning Program