MCP๋กœ ์—ฐ๊ฒฐ โ†’

๊ณ„์‚ฐ ์ž…๋ ฅ

๊ณต์‹

๊ณต์‹: ๋ฐ˜ํŠธํ˜ธํ”„ ๋ฐฉ์ •์‹ ๊ณ„์‚ฐ๊ธฐ
Show calculation steps (1)
  1. Solve for K2

    Solve for K2: ๋ฐ˜ํŠธํ˜ธํ”„ ๋ฐฉ์ •์‹ ๊ณ„์‚ฐ๊ธฐ

    Exponentiate to obtain the equilibrium constant at the second temperature.

๊ด‘๊ณ 

๊ฒฐ๊ณผ

T2์—์„œ์˜ ํ‰ํ˜• ์ƒ์ˆ˜ (K2)
1.925583
๋ฌด์ฐจ์›
ln(K2/K1) 0.655229
K2 / K1 ๋น„์œจ 1.925583

๋ฐ˜ํŠธํ˜ธํ”„ ๋ฐฉ์ •์‹์ด๋ž€?

๋ฐ˜ํŠธํ˜ธํ”„ ๋ฐฉ์ •์‹(Van't Hoff equation)์€ ํ™”ํ•™ ๋ฐ˜์‘์˜ ํ‰ํ˜• ์ƒ์ˆ˜ K๊ฐ€ ์˜จ๋„์— ๋”ฐ๋ผ ์–ด๋–ป๊ฒŒ ๋ณ€ํ•˜๋Š”์ง€๋ฅผ ๋‚˜ํƒ€๋ƒ…๋‹ˆ๋‹ค. ๋‘ ์˜จ๋„์—์„œ์˜ ํ‰ํ˜• ์ƒ์ˆ˜ ๋น„์œจ์„ ํ‘œ์ค€ ๋ฐ˜์‘ ์—”ํƒˆํ”ผ ฮ”H์™€ ์—ฐ๊ฒฐํ•ด ์ฃผ๋Š” ์‹์ด์ฃ . ์ด ๊ณ„์‚ฐ๊ธฐ๋Š” ์ธก์ •ํ•œ ์˜จ๋„ ๊ตฌ๊ฐ„์—์„œ ฮ”H๊ฐ€ ์ผ์ •ํ•˜๋‹ค๊ณ  ๊ฐ€์ •ํ•˜๋Š” 2์  ์ ๋ถ„ํ˜• ์‹์„ ์‚ฌ์šฉํ•˜๋ฉฐ, ์ƒˆ ํ‰ํ˜• ์ƒ์ˆ˜ K2์™€ ํ•จ๊ป˜ \(\ln(K_2/K_1)\) ๊ฐ’, ๊ทธ๋ฆฌ๊ณ  \(K_2/K_1\) ๋น„์œจ์„ ํ•จ๊ป˜ ์•Œ๋ ค ์ค๋‹ˆ๋‹ค.

๋‘ ์ ์ด ํ‘œ์‹œ๋œ ln K ๋Œ€ 1/T์˜ ์ง์„  ๊ทธ๋ž˜ํ”„
๋ฐ˜ํŠธํ˜ธํ”„ ๋ฐฉ์ •์‹์€ ln K์™€ 1/T ์‚ฌ์ด์˜ ์„ ํ˜• ๊ด€๊ณ„๋ฅผ ๋‚˜ํƒ€๋ƒ…๋‹ˆ๋‹ค.

์‚ฌ์šฉ ๋ฐฉ๋ฒ•

์˜จ๋„ T1(์ผˆ๋นˆ ๋‹จ์œ„)์—์„œ ์ด๋ฏธ ์•Œ๊ณ  ์žˆ๋Š” ํ‰ํ˜• ์ƒ์ˆ˜ K1, ํ‘œ์ค€ ์—”ํƒˆํ”ผ ๋ณ€ํ™” ฮ”H(kJ/mol), ๊ทธ๋ฆฌ๊ณ  ๋‘ ๋ฒˆ์งธ ์˜จ๋„ T2(์ผˆ๋นˆ ๋‹จ์œ„)๋ฅผ ์ž…๋ ฅํ•˜์„ธ์š”. ๊ณ„์‚ฐ๊ธฐ๋Š” ฮ”H๋ฅผ ์ค„(J) ๋‹จ์œ„๋กœ ๋ณ€ํ™˜ํ•˜๊ณ  ๊ธฐ์ฒด ์ƒ์ˆ˜ \(R = 8.314\ \text{J/mol}\cdot\text{K}\)๋ฅผ ์ ์šฉํ•ด K2๋ฅผ ๊ณ„์‚ฐํ•ฉ๋‹ˆ๋‹ค. ฮ”H๊ฐ€ ์–‘์ˆ˜(ํก์—ด ๋ฐ˜์‘)์ด๋ฉด ์˜จ๋„๊ฐ€ ์˜ฌ๋ผ๊ฐˆ์ˆ˜๋ก K๊ฐ€ ์ปค์ง€๊ณ , ฮ”H๊ฐ€ ์Œ์ˆ˜(๋ฐœ์—ด ๋ฐ˜์‘)์ด๋ฉด ์˜จ๋„๊ฐ€ ์˜ฌ๋ผ๊ฐˆ์ˆ˜๋ก K๊ฐ€ ์ž‘์•„์ง‘๋‹ˆ๋‹ค.

๊ณต์‹ ํ’€์ด

์ ๋ถ„ํ˜• ์‹์€ ๋‹ค์Œ๊ณผ ๊ฐ™์Šต๋‹ˆ๋‹ค.

$$\ln\!\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$

์ด๋ฅผ K2์— ๋Œ€ํ•ด ํ’€๋ฉด ๋‹ค์Œ์ด ๋ฉ๋‹ˆ๋‹ค.

$$K_2 = K_1\,\exp\!\left[-\frac{\Delta H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\right]$$

์˜จ๋„๋Š” ๋ฐ˜๋“œ์‹œ ์ผˆ๋นˆ ๋‹จ์œ„์—ฌ์•ผ ํ•˜๋ฉฐ, ฮ”H์™€ R์€ ๋™์ผํ•œ ์—๋„ˆ์ง€ ๋‹จ์œ„๋ฅผ ์จ์•ผ ํ•˜๊ธฐ ๋•Œ๋ฌธ์— ๊ณ„์‚ฐ ๊ณผ์ •์—์„œ ฮ”H๋ฅผ kJ/mol์—์„œ J/mol๋กœ ์ž๋™ ๋ณ€ํ™˜ํ•ฉ๋‹ˆ๋‹ค.

๊ณ„์‚ฐ ์˜ˆ์‹œ

K1 = 1, T1 = 298 K, ฮ”H = 50 kJ/mol, T2 = 308 K๋ผ๊ณ  ๊ฐ€์ •ํ•ด ๋ด…์‹œ๋‹ค. ์ด๋•Œ ๋‹ค์Œ๊ณผ ๊ฐ™์Šต๋‹ˆ๋‹ค.

$$\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{308} - \frac{1}{298} = -0.00010897\ \text{K}^{-1}$$

\(\Delta H/R = 50000/8.314 = 6013.95\)์ด๋ฏ€๋กœ

$$\ln\!\left(\frac{K_2}{K_1}\right) = -(6013.95)(-0.00010897) = 0.65535$$

์ด๊ณ , \(K_2 = 1 \cdot e^{0.65535} \approx 1.9258\)์ด ๋ฉ๋‹ˆ๋‹ค. ์ด ํก์—ด ๋ฐ˜์‘์—์„œ๋Š” ์˜จ๋„๊ฐ€ 10 K๋งŒ ์˜ฌ๋ผ๊ฐ€๋„ ํ‰ํ˜• ์ƒ์ˆ˜๊ฐ€ ๊ฑฐ์˜ ๋‘ ๋ฐฐ๋กœ ์ปค์ง€๋Š” ์…ˆ์ž…๋‹ˆ๋‹ค.

์ž์ฃผ ๋ฌป๋Š” ์งˆ๋ฌธ

์˜จ๋„๋Š” ๊ผญ ์ผˆ๋นˆ ๋‹จ์œ„๋กœ ๋„ฃ์–ด์•ผ ํ•˜๋‚˜์š”? ๋„ค. ์ด ๋ฐฉ์ •์‹์€ ์ ˆ๋Œ€ ์˜จ๋„๋ฅผ ์š”๊ตฌํ•˜๋ฏ€๋กœ, ์„ญ์”จ(ยฐC) ๊ฐ’์— 273.15๋ฅผ ๋”ํ•ด ์ผˆ๋นˆ์œผ๋กœ ๋ณ€ํ™˜ํ•˜์„ธ์š”.

R ๊ฐ’์€ ๋ฌด์—‡์„ ์“ฐ๋‚˜์š”? \(R = 8.314\ \text{J/(mol}\cdot\text{K)}\)๋ฅผ ์‚ฌ์šฉํ•˜๋ฉฐ, ฮ”H๋ฅผ kJ/mol์—์„œ J/mol๋กœ ๋ณ€ํ™˜ํ•ด ๋‹จ์œ„๊ฐ€ ์„œ๋กœ ์ƒ์‡„๋˜๋„๋ก ํ•ฉ๋‹ˆ๋‹ค.

ฮ”H๋Š” ์ผ์ •ํ•˜๋‹ค๊ณ  ๊ฐ€์ •ํ•˜๋‚˜์š”? ๋„ค. 2์  ์‹์€ ํ•ด๋‹น ์˜จ๋„ ๊ตฌ๊ฐ„์—์„œ ฮ”H(๊ทธ๋ฆฌ๊ณ  ฮ”S)๊ฐ€ ํฌ๊ฒŒ ๋ณ€ํ•˜์ง€ ์•Š๋Š”๋‹ค๊ณ  ๊ฐ€์ •ํ•ฉ๋‹ˆ๋‹ค. ์˜จ๋„ ๋ณ€ํ™” ํญ์ด ํฌ์ง€ ์•Š์„ ๋•Œ๋Š” ์ถฉ๋ถ„ํžˆ ์ข‹์€ ๊ทผ์‚ฌ๊ฐ€ ๋ฉ๋‹ˆ๋‹ค.

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