Chemical and Ionic Equilibrium - Result Question 129
####8. $5.1 g NH _4 SH$ is introduced in $3.0 L$ evacuated flask at $327^{\circ} C .30 %$ of the solid $NH _4 SH$ decomposed to $NH _3$ and $H _2 S$ as gases. The $K _p$ of the reaction at $327^{\circ} C$ is $\left(R=0.082 atm mol^{-1} K^{-1}\right.$, molar mass of $S=32 g mol^{-1}$, molar mass of $N=14 g mol^{-1}$ )
(2019 Main, 10 Jan II)
(a) $0.242 \times 10^{-4} atm^{2}$
(b) $0.242 atm^{2}$
(c) $4.9 \times 10^{-3} atm^{2}$
(d) $1 \times 10^{-4} atm^{2}$
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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. $SO _4^{2-}$ | 46. $F$ | 47. $F$ |
| 48. $F$ | 49. $(3)$ | 50. $\left(1.6 \times 10^{-7}\right)$ | 52. $(8)$ |
| 53. $(9)$ | 55. $(4.86)$ | 56. $\left(1.2 \times 10^{-3} M\right)$ | |
| 58. $\left(2 \times 10^{-8}\right)$ | 61. $(11.5)$ | 62. $(6.50)$ | 64. (1) |
| 65. $(80)$ | 67. $\left(1.8 \times 10^{-5}\right)$ | 68. $\left(9.67 \times 10^{-11}\right)$ | 69. |
| $\left(27.78 \times 10^{3}\right)$ | 74. $(4.20)$ | ||
| 71. $(0.177)$ | 72. $\left(8.7 \times 10^{-4} gL^{-1}\right)$ | ||
| 75. $(99.83)$ | 77. $(>7)$ |
Solution:
- Molar mass of $NH _4 SH=18+33=51 g mol^{-1}$
Number of moles of $NH _4 SH$ introduced in the vessel
$$ \begin{aligned} & =\frac{\text { Weight }}{\text { Molar mass }}=\frac{5 \cdot 1}{51}=0.1 mol \ & NH _4 SH(s) \rightleftharpoons NH _3(g)+H _2 S(g) \ & \text { Number of } \quad 0.1 \quad 0 \quad 0 \ & \text { moles at } t=0 \ & \text { At } t=t _{\text {eq }} \quad 0.1(1-0.03) \quad 30 % \text { of } \quad 30 % \text { of } 0.1 \ & \begin{array}{lll} \begin{array}{l} \text { Active mass } \ \left(mol L^{-1}\right) \end{array} & \frac{0.03}{3}=0.01 & \frac{0.03}{3}=0.01 \end{array} \ & K _C=\frac{\left[NH _3\right]\left[H _2 S\right]}{\left[NH _4 HS(s)\right]}=\frac{0.01 \times 0.01}{1}=10^{-4}\left(mol L^{-1}\right)^{2} \ & \Rightarrow \quad K _p=K _C(R T)^{\Delta n _g} \ & \text { [where, } \left.\Delta n _g=\Sigma n _{\text {product }}-\Sigma n _{\text {reactant }}\right]=2-0=2 \ & \therefore \quad K _p=K _C(R T)^{2} \ & =10^{-4} \times[0.082 \times(273+327)]^{2} atm^{2} \ & =0.242 atm^{2} \end{aligned} $$
