Bank Soal Kimia SMA Hukum, Tetapan, dan Pergeseran Kesetimbangan

Diketahui:

• Persamaan reaksi: CH₄(g) + 2O₂(g) ⇌ CO₂(g) + 2H₂O(g)
• n awal CH₄ = 1 mol
• n awal O₂ = 2 mol
• V = 2 L
• n CH₄ saat setimbang = 0,75 mol

Ditanyakan:

$K_{\text{}\text{p}}?\ \$

Dijawab:

K_p ⟶ tetapan kesetimbangan berdasarkan tekanan parsial.

Secara umum untuk reaksi kesetimbangan berikut:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

rumusan tetapan kesetimbangan berdasarkan tekanan (K_p​) adalah 

$K_{\text{p}}=\frac{\left(P_{\text{C}}^{ }\right)^c\left(P_{\text{D}}\right)^d}{\left(P_{\text{A}}\right)^a\left(P_{\text{B}}\right)^b}\$

Keterangan:

• 𝑃_𝐴 = Tekanan parsial zat A saat setimbang
• 𝑃_𝐵 = Tekanan parsial zat B saat setimbang
• 𝑃_C = Tekanan parsial zat C sat setimbang
• 𝑃_𝐷 = Tekanan parsial zat D saat setimbang 

*Hanya fasa gas (g) yang disertakan dalam persamaan tetapan kesetimbangan berdasarkan tekanan.

Untuk reaksi:

CH₄(g) + 2O₂(g) ⇌ CO₂(g) + 2H₂O(g)

Tetapan kesetimbangannya adalah

$K_{\text{p}}=\frac{\left(P_{\text{CO}_2}^{ }\right)^{ }\left(P_{\text{H}_2\text{O}}\right)^2}{\left(P_{\text{CH}_4}^{ }\right)^{ }\left(P_{\text{O}_2\text{}}\right)^2}\ \$

• Buat persamaan reaksi kesetimbangan dan tentukan mol mula-mula, mol reaksi, dan mol saat setimbang

n total = mol pereaksi pada saat setimbang + mol produk pada saat setimbang

n total = mol CH₄ + mol O₂ + mol CO₂ + mol H₂O

n total = 0,75 mol + 1,5 mol + 0,25 mol + 0,5 mol

n total = 3 mol

• Tentukan tekanan parsial masing-masing zat

$P=\frac{\text{mol gas yang ditanyakan}}{\text{mol gas total}}\times P_{\text{total}}$



$P_{\text{CH}_4}=\frac{\text{mol CH}_4}{\text{mol gas total}}\times P_{\text{total}}$

$P_{\text{CH}_4}=\frac{0,75\ \text{mol }}{\text{3 mol}}\times5\ \text{atm}$

$P_{\text{CH}_4}=\frac{1}{\text{4}}\times5\ \text{atm}$

$P_{\text{CH}_4}=\frac{5}{\text{4}}\ \text{atm}\$

$P_{\text{O}_2}=\frac{\text{mol O}_2}{\text{mol gas total}}\times P_{\text{total}}\$

$P_{\text{O}_2}=\frac{1,5\ \text{mol }}{\text{3 mol}}\times5\ \text{atm}\ \$

$P_{\text{O}_2}=\frac{1}{2}\text{}\times5\ \text{atm}\ \ \$

$P_{\text{O}_2}=\frac{5}{2}\text{}\ \text{atm}\ \ \ \$

$P_{\text{CO}_2}=\frac{\text{mol CO}_2}{\text{mol gas total}}\times P_{\text{total}}\$

$P_{\text{CO}_2}=\frac{0,25\ \text{mol }}{\text{3 mol}}\times5\ \text{atm}$

$P_{\text{CO}_2}=\frac{1}{\text{12}}\times5\ \text{atm}\$

$P_{\text{CO}_2}=\frac{5}{\text{12}}\ \text{atm}\ \$

$P_{\text{H}_2\text{O}}=\frac{\text{mol }\text{H}_2\text{O}}{\text{mol gas total}}\times P_{\text{total}}\$

$P_{\text{H}_2\text{O}}=\frac{0,5\ \text{mol }}{\text{3 mol}}\times5\ \text{atm}\$

$P_{\text{H}_2\text{O}}=\frac{1}{\text{6}}\times5\ \text{atm}\$

$P_{\text{H}_2\text{O}}=\frac{5}{\text{6}}\ \text{atm}\ \$

• Hitung tetapan kesetimbangan

$K_{\text{p}}=\frac{\left(P_{\text{CO}_2}^{ }\right)^{ }\left(P_{\text{H}_2\text{O}}\right)^2}{\left(P_{\text{CH}_4}^{ }\right)^{ }\left(P_{\text{O}_2\text{}}\right)^2}\ \ \$

$K_{\text{p}}=\frac{\left(\frac{5}{12}\right)\left(\frac{5}{6}\right)^2}{\left(\frac{5}{4}\right)\left(\frac{5}{2}\text{}\right)^{^2}}\ \ \ \$

$K_{\text{p}}=\frac{\left(\frac{5}{12}\right)\left(\frac{25}{36}\right)^{ }}{\left(\frac{5}{4}\right)\left(\frac{25}{4}\text{}\right)}\ \ \ \ \$

$K_{\text{p}}=\frac{\left(\frac{1}{12}\right)\left(\frac{1}{36}\right)}{\left(\frac{1}{4}\right)\left(\frac{1}{4}\text{}\right)}\ \ \ \ \$

$K_{\text{p}}=\frac{4\ \times\ 4}{12\ \times\ 36}$

$K_{\text{p}}=\frac{1}{27}\ \ \ \ \$

Jadi K_p untuk soal di atas adalah $\frac{1}{27}$.