SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 129

####8. $5.1 g N H_{4} S H$ is introduced in $3.0 L$ evacuated flask at $327^{\circ} C .30$ of the solid $N H_{4} S H$ decomposed to $N H_{3}$ and $H_{2} S$ as gases. The $K_{p}$ of the reaction at $327^{\circ} C$ is $(R = 0.082 a t m m o l^{- 1} K^{- 1}$ , molar mass of $S = 32 g m o l^{- 1}$ , molar mass of $N = 14 g m o l^{- 1}$ )

(2019 Main, 10 Jan II)

(a) $0.242 \times 10^{- 4} a t m^{2}$

(b) $0.242 a t m^{2}$

(d) $1 \times 10^{- 4} a t m^{2}$

Show Answer

Answer:

Correct Answer: 8. (c)

9. $(d)$	10. $(b)$	11. (b)	12. (d)
13. $(d)$	14. (d)	15. (b)	16. (a)
17. (c)	18. (d)	19. (a)	20. (b)
21. (d)	22. (d)	23. (a)	24. (a)
25. (d)	26. (c)	27. (a)	28. (d)
29. (c)	30. (a)	31. (b)	32. (a)
33. (a)	34. (b)	35. (d)	36. (b)
37. (c, d)	38. $(a, b, c)$	39. (b, c)	40. (4.47)
41. $(d)$	42. $I_{2}$	43. hydration
44. amphoteric	45. $S O_{4}^{2 -}$	46. $F$	47. $F$
48. $F$	49. $(3)$	50. $(1.6 \times 10^{- 7})$	52. $(8)$
53. $(9)$	55. $(4.86)$	56. $(1.2 \times 10^{- 3} M)$
58. $(2 \times 10^{- 8})$	61. $(11.5)$	62. $(6.50)$	64. (1)
65. $(80)$	67. $(1.8 \times 10^{- 5})$	68. $(9.67 \times 10^{- 11})$	69.
$(27.78 \times 10^{3})$			74. $(4.20)$
71. $(0.177)$	72. $(8.7 \times 10^{- 4} g L^{- 1})$
75. $(99.83)$	77. $(> 7)$

Solution:

Molar mass of $N H_{4} S H = 18 + 33 = 51 g m o l^{- 1}$

Number of moles of $N H_{4} S H$ introduced in the vessel

$\begin{aligned} = \frac{Weight}{Molar mass} = \frac{5 \cdot 1}{51} = 0.1 m o l & N H_{4} S H (s) ⇌ N H_{3} (g) + H_{2} S (g) & Number of 0.1 0 0 & moles at t = 0 & At t = t_{eq} 0.1 (1 - 0.03) 30 & \begin{array}{lll} \begin{matrix} Active mass (m o l L^{- 1}) \end{matrix} & \frac{0.03}{3} = 0.01 & \frac{0.03}{3} = 0.01 \end{array} & K_{C} = \frac{[N H_{3}] [H_{2} S]}{[N H_{4} H S (s)]} = \frac{0.01 \times 0.01}{1} = 10^{- 4} {(m o l L^{- 1})}^{2} & \Rightarrow K_{p} = K_{C} (R T)^{Δ n_{g}} & [where, Δ n_{g} = Σ n_{product} - Σ n_{reactant}] = 2 - 0 = 2 & ∴ K_{p} = K_{C} (R T)^{2} & = 10^{- 4} \times [0.082 \times (273 + 327)]^{2} a t m^{2} & = 0.242 a t m^{2} \end{aligned}$