AI-তৈরি নমুনা উত্তর
বিক্রিয়াটি হলো:1 2 N 2 ( g ) + 3 2 H 2 ( g ) ⇌ NH 3 ( g ) \frac{1}{2}\text{N}_2(\text{g}) + \frac{3}{2}\text{H}_2(\text{g}) \rightleftharpoons \text{NH}_3(\text{g}) 2 1 N 2 ( g ) + 2 3 H 2 ( g ) ⇌ NH 3 ( g )
মোট চাপ, P = 20 atm P = 20\text{ atm} P = 20 atm NH 3 \text{NH}_3 NH 3 = 40 % = 40\% = 40% NH 3 \text{NH}_3 NH 3 X NH 3 = 0.40 X_{\text{NH}_3} = 0.40 X NH 3 = 0.40 NH 3 \text{NH}_3 NH 3 p NH 3 = 0.40 × 20 atm = 8 atm p_{\text{NH}_3} = 0.40 \times 20\text{ atm} = 8\text{ atm} p NH 3 = 0.40 × 20 atm = 8 atm
অবশিষ্ট গ্যাসদ্বয়ের (N 2 \text{N}_2 N 2 H 2 \text{H}_2 H 2 = 100 % − 40 % = 60 % = 100\% - 40\% = 60\% = 100% − 40% = 60% N 2 : H 2 = 1 : 3 \text{N}_2 : \text{H}_2 = 1 : 3 N 2 : H 2 = 1 : 3 N 2 \text{N}_2 N 2 X N 2 = 0.60 × 1 4 = 0.15 X_{\text{N}_2} = 0.60 \times \frac{1}{4} = 0.15 X N 2 = 0.60 × 4 1 = 0.15 H 2 \text{H}_2 H 2 X H 2 = 0.60 × 3 4 = 0.45 X_{\text{H}_2} = 0.60 \times \frac{3}{4} = 0.45 X H 2 = 0.60 × 4 3 = 0.45
আংশিক চাপসমূহ:p N 2 = 0.15 × 20 atm = 3 atm p_{\text{N}_2} = 0.15 \times 20\text{ atm} = 3\text{ atm} p N 2 = 0.15 × 20 atm = 3 atm p H 2 = 0.45 × 20 atm = 9 atm p_{\text{H}_2} = 0.45 \times 20\text{ atm} = 9\text{ atm} p H 2 = 0.45 × 20 atm = 9 atm
সাম্যধ্রুবক, K p K_p K p K p = p NH 3 ( p N 2 ) 1 / 2 ⋅ ( p H 2 ) 3 / 2 K_p = \frac{p_{\text{NH}_3}}{(p_{\text{N}_2})^{1/2} \cdot (p_{\text{H}_2})^{3/2}} K p = ( p N 2 ) 1/2 ⋅ ( p H 2 ) 3/2 p NH 3 K p = 8 ( 3 ) 1 / 2 ⋅ ( 9 ) 3 / 2 = 8 3 ⋅ 27 = 8 27 3 ≈ 0.171 atm − 1 K_p = \frac{8}{(3)^{1/2} \cdot (9)^{3/2}} = \frac{8}{\sqrt{3} \cdot 27} = \frac{8}{27\sqrt{3}} \approx 0.171\text{ atm}^{-1} K p = ( 3 ) 1/2 ⋅ ( 9 ) 3/2 8 = 3 ⋅ 27 8 = 27 3 8 ≈ 0.171 atm − 1