Bank Soal Kimia SMA Hukum, Tetapan, dan Pergeseran Kesetimbangan

Diketahui:

mol NH₃ setimbang = 0,6 mol

mol HCl setimbang = 0,5 mol

mol NH₄Cl setimbang = 0,9 mol

volume = 3 liter

tekanan gas = 32,8 atm

Ditanya:

Berapa suhu ruang?

Jawab:

Mencari nilai K_c:

$\left[\text{NH}_3\right]=\frac{0,6\ \text{mol}}{3\ \text{liter}}=0,2\ \text{M}$

$\left[\text{HCl}\right]=\frac{0,5\ \text{mol}}{3\ \text{liter}}=0,16\ \text{M}$

$\left[\text{NH}_4\text{Cl}\right]=\frac{0,9\ \text{mol}}{3\ \text{liter}}=0,3\ \text{M}$

$K_{\text{c}}=\frac{\left[\text{NH}_4\text{Cl}\right]}{\left[\text{NH}_3\right]\ \left[\text{HCl}\right]}$

$=\frac{0,3}{0,2\ \times\ 0,16}$

$=9,375\ \frac{\text{liter}}{\text{mol}}$

Mencari nilai K_p:

$P_{\text{NH}_3}=\frac{\text{mol}\ \text{NH}_3}{\text{mol total}}\times P_{\text{total}}=\frac{0,6}{2\ }\times32,8\ \text{atm}=9,84\ \text{atm}$

$P_{\text{HCl}}=\frac{\text{mol}\ \text{HCl}}{\text{mol total}}\times P_{\text{total}}=\frac{0,5}{2\ }\times32,8\ \text{atm}=8,2\ \text{atm}$

$P_{\text{NH}_4\text{Cl}}=\frac{\text{mol}\ \text{NH}_4\text{Cl}}{\text{mol total}}\times P_{\text{total}}=\frac{0,9}{2\ }\times32,8\ \text{atm}=14,76\ \text{atm}$

$K_{\text{p}}=\frac{\left(P_{\text{NH}_4\text{Cl}}\right)}{\left(P_{\text{NH}_3}\right)\ \left(P_{\text{HCl}}\right)}$

$=\frac{14,76}{9,84\times8,2}$

$=0,182$ atm^-1

Mencari suhu ruang:

NH_3($g$) + HCl_($g$) ⇌ NH₄Cl_($g$)

$K_{\text{p}}=K_{\text{c}}\times\left(RT\right)^{∆n}$ ; ∆$n$ = koef produk - koef reaktan = 1 - (1 + 1) = -1

$K_{\text{p}}=K_{\text{c}}\times\left(RT\right)^{-1}$

$K_{\text{p}}=\frac{K_{\text{c}}}{RT}$

$0,182\ \text{atm}^{-1}=\frac{9,375\ \frac{\text{liter}}{\text{mol}}}{0,082\ \frac{\text{liter.atm}}{\text{mol.K}}\times T}$

$T=628\ \text{K}\ =\ 355\ ^{\text{o}}\text{C}$

Jadi, suhu ruang tersebut adalah 355 ^oC.