An experiment is run and found to obey the rate law rate = kX 2 Y. Additionally,

Question

An experiment is run and found to obey the rate law rate = kX2Y.  Additionally, it is found that it has an activation energy of 167,943 J/mol. If the rate constant for the reaction is 2.5 x 10-2 M-2s-1at 224 oC, what is the rate constant at 215 oC? An experiment is run and found to obey the rate law rate = kX2Y.  Additionally, it is found that it has an activation energy of 167,943 J/mol. If the rate constant for the reaction is 2.5 x 10-2 M-2s-1at 224 oC, what is the rate constant at 215 oC? An experiment is run and found to obey the rate law rate = kX2Y.  Additionally, it is found that it has an activation energy of 167,943 J/mol. If the rate constant for the reaction is 2.5 x 10-2 M-2s-1at 224 oC, what is the rate constant at 215 oC?

Explanation / Answer

log[k2/k1]=Ea/2.303R[1/T1-1/T2]      T2>T1

T2= 497 K = 224 C

T1=488 K = 215 C

K2= 2.5*10^-2 M-2.S-1

log[2.5*10^-2/K1]= 167943/2.303*8.314[1/488-1/497]

on simplification

K1= 1.18*10^-2 M^-2. s^-1

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