Temp 20 Firstorder Reaction Graph FirstOrder Linear Regression

Temp. = 20 ℃ <First-order Reaction Graph>

First-Order < Linear Regression >

Temp. = 20 ℃ Second-order Reaction Graph

Second-Order Reaction : Result ①

Second-Order Reaction : Result ② � Temp. = 20 ℃ / Second-Order Reaction 1/A 1 = kt + 1/A 0 반응상수 [kt=20, 2차] 값 = 3. 422 ( + 부호 ) 기여율 r 2 값 = 0. 913 ※ 기여율 r 2 값 : 0. 913 < 0. 95 신뢰성을 못 갖는다

Temp. = 20 ℃<Third-order Reaction Graph>

Third-Order Reaction : Result ①

Third-Order Reaction : Result ② � Temp . = 20 ℃ / Third-Order Reaction 1/2 A 12 = kt + 1/2 A 02 반응상수 [kt=20, 3차] 값 = 3. 877 ( + 부호 ) 기여율 r 2 값 = 0. 783 ※ 기여율 r 2 값 : 0. 783 < 0. 95 신뢰성을 못 갖는다.

Temp. = 30 ℃<zero-order Reaction Graph>

Zero-Order : Linear Regression

Zero-Order Reaction : Result ② � Temp . = 30 ℃ / Zero-Order Reaction A 2 = - kt + A 0 반응상수 [kt=30, 0차] 값 = 0. 473 ( - 부호 ) 기여율 r 2 값 = 0. 977 ※ 기여율 r 2 값 : 0. 977 > 0. 95 신뢰성을 갖는다.

Temp. = 30 ℃<First-order Reaction Graph>

First-Order : Linear Regression

First-Order Reaction : Result � Temp . = 30 ℃ / First-Order Reaction ln. A 2 = - kt + ln. A 0 반응상수 [kt=30, 1차] 값 = 0. 071 ( - 부호 ) 기여율 r 2 값 = 0. 937 ※ 기여율 r 2 값 : 0. 937 < 0. 95 다 신뢰성을 못 갖는

Temp. = 30 ℃<Second-order Reaction Graph>

Second-Order Reaction : Result ①

Temp. = 30 ℃<Third-order Reaction Graph>

Third-Order Reaction : Result ①

Third-Order Reaction : Result ② � Temp 호) . = 30 ℃ / Third-Order Reaction 1/2 A 22 = kt + 1/2 A 02 반응상수 [kt=30, 3차] 값 = 8. 101 ( + 부 기여율 r 2 값 = 0. 460 ※ 기여율 r 2 값 : 0. 460 < 0. 95 못 갖는다 신뢰성을

Eact 계산 ② ┌ t 1 = 20 ℃ = 293. 15 k │ t 2 = 30 ℃ = 303. 15 k └ R = 8. 314 J/molㆍK / k 1 = 3. 422 ( Second-Order Reaction ) / k 2 = 0. 473 ( Zero-Order Reaction ) Ea = 8. 314 J/molㆍK * {[293. 15 K * 303. 15 K]/[303. 15 K – 293. 15 K]} * ln[0. 473/3. 422] = - 146. 2 KJ / mol