Question
Sun January 13, 2013

# see in description

Sun January 13, 2013

The Arrhenius equation is

log10 k2/k1 = Ea/(RÃ—2.303) [(T2-T1)/(T1 T2 )]

Given:  k2/k1 = 3; R=8.314 JK-1 molÂ–1; T1 = 20 + 273 = 293 K

and   T2 = 50 + 273 = 323 K

Subtracting the given values in the Arrhenius equation,

log10 3 = Ea/(8.314Ã—2.303) [(323-293)/(323Ã—293)]

Ea = (2.303 Ã— 8.314Ã—323 Ã— 293Ã—0.477)/30

= 28811.8 J mol-1

= 28.8118 kJ mol-1

Related Questions
Sat September 23, 2017

# for the reaction A+B____PRODUCT,following initial rates were obtained at various given initial concentrations  S.No         [A]                  [B]                          Rate molL-1S-1 1              0.1                  0.1                           0.05 2              0.2                   0.1                             0.10 3               0.1                  0.2                            0.05 Write the rate law and find the rate constant for the above data. {PLEASE EXPLAIN STEPWISE}

Mon September 18, 2017