Answer to Question #143989 in General Chemistry for gia

Question #143989
The rate constant of a first-order reaction is 3.75
x 10-4
s
-1 at 300°C. If the activation energy is 101
kJ/mol, calculate the temperature at which its rate
constant is 7.50 x 10-4
s
-1
.
1
Expert's answer
2020-11-14T13:55:50-0500
k=AeEaRTk = Ae^{ { -E_a \over RT}}

ln(k2k1)=Ea(T2T1)RT2T1ln({k_2 \over k_1}) = {E_a(T_2-T_1) \over RT_2T_1}

ln(7.501043.75104)=101000(x573)8.314573xln({7.50*10^{-4} \over 3.75*10^{-4}}) = {101 000(x-573) \over 8.314*573x}

0.69=101000x578730004764x0.69 = {101000x - 57873000 \over 4764x}

3287.16x=101000x578730003287.16x = 101000x - 57873000

57873000=97712.84x57873000 = 97712.84x

x=592.27K=592Kx = 592.27 K = 592K

T=592K=3190CT = 592K = 319^0C


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