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A first order reaction has a rate constant of 0.981 sec-1 at 26.7 oC. If the activation energy is 61 kJ, calculate the temperature in oC at which the rate constant is 0.425 sec-1. Use 1 decimal place

Respuesta :

Answer:

T2 = 289.8 K

Explanation:

K1 =  0.981 sec-1

T1 = 26.7 oC = 299.7 K (Converting to Kelvin temperature)

Ea = 61kJ = 61,000 J

K2 = 0.425 sec-1

T2 = ?

The equation relating these parameters is;

ln(K2 / K) = - Ea/R (1 / T2 - 1 / T1)

R is the gas constant (8.314 J/mol-K)

ln (0.425 / 0.981) = - 61,000 / 8.314 (1 / T2 - 1 / T1 )

ln (0.4332) = -7337.02 (1 / T2 - 1 / T1 )

- 0.83656 = - 7337.02 (1 / T2 - 1 / T1 )

0.0001140 = ( 1 / T2 - 1 / 299.7 )

0.0001140 = 1 / T2 - 0.00333667

1 / T2 = 0.0001140 + 0.00333667

1 / T2 = 0.00345067

T2 = 289.8 K

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