contestada

The rate constant of the elementary reaction C2H5CN(g) → CH2CHCN(g) + H2(g) is k = 7.21×10-3 s-1 at 634 °C, and the reaction has an activation energy of 247 kJ/mol. (a) Compute the rate constant of the reaction at a temperature of 741 °C. _________ s-1 (b) At a temperature of 634 °C, 96.1 s is required for half of the C2H5CN originally present to be consume. How long will it take to consume half of the reactant if an identical experiment is performed at 741 °C? (Enter numbers as numbers, no units. For example, 300 minutes would be 300. For letters, enter A, B, or C. Enter numbers in scientific notation using e# format. For example 1.43×10-4 would be 1.43e-4.)

Respuesta :

ajeigbeibraheem

Answer:

the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]

it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C

Explanation:

Given that :

[tex]k_1 = 7.21*10^{-3} s^{-1} \\ \\ E_a = 247 kJ/mol \ \ \ = 247*10^3 \ J/mol \\ \\ T_1 = 634 ^ {^0} C= (273 + 634) K = 907 \ K \\ \\ T_2 = 741^{^0 } C = (273+ 741) K = 1014 \ K \\ \\ R =8.314 \ \ J/mol/K[/tex]

a)

According to Arrhenius Equation ;

[tex]In\frac{k_2}{k_1} = -\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]

[tex]In\frac{k_2}{7.21*10^{-3}} = 3.456412[/tex]

[tex]\frac{k_2}{7.21*10^{-3}} =e^{ 3.456412[/tex]

[tex]k_2 = 31.70302 * 7.21*10^{-3}[/tex]

[tex]k_2 = 0.22858 \ s^{-1}[/tex]

Therefore ,  the rate constant of the reaction at a temperature of 741 °C is [tex]0.22858 \ s^{-1}[/tex]

b) Given that  :

[tex]t_{(1/2)} = 96.1 \ s[/tex]

[tex]k_1 = \frac{0.693}{ t_{(1/2)}}[/tex]

[tex]k_1 = \frac{0.693}{ 96.1 \ s}[/tex]

[tex]k_1 = 7.211*10^{-3} \ s^{-1}[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{1}{1014} -\frac{1}{907} )[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = -\frac{247*10^3}{8.314}(\frac{907-1014}{907*1014} )[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = \frac{-247*10^3*-107}{8.314*907*1014}[/tex]

[tex]In\frac{k_2}{7.211*10^{-3}} = 3.456412[/tex]

[tex]\frac{k_2}{7.211*10^{-3}} =e^{ 3.456412[/tex]

[tex]k_2 =e^{ 3.456412 * 7.21*10^{-3}[/tex]

[tex]k_2 = 0.22862 \ s^{-1}[/tex]

[tex]t_{(1/2)_2}} = \frac{0.693}{k_2}[/tex]

[tex]t_{(1/2)_2}} = \frac{0.693}{0.22862}[/tex]

[tex]t_{(1/2)_2}} =3.0313 \ s[/tex]

it will take 3.0313 s to consume half of the reactant if an identical experiment is performed at 741 °C

Otras preguntas