Answer to Question #143989 in General Chemistry for gia
2020-11-11T18:59:40-05:00
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
2020-11-14T13:55:50-0500
k = A e − E a R T k = Ae^{ { -E_a \over RT}} k = A e RT − E a
l n ( k 2 k 1 ) = E a ( T 2 − T 1 ) R T 2 T 1 ln({k_2 \over k_1}) = {E_a(T_2-T_1) \over RT_2T_1} l n ( k 1 k 2 ) = R T 2 T 1 E a ( T 2 − T 1 )
l n ( 7.50 ∗ 1 0 − 4 3.75 ∗ 1 0 − 4 ) = 101000 ( x − 573 ) 8.314 ∗ 573 x ln({7.50*10^{-4} \over 3.75*10^{-4}}) = {101 000(x-573) \over 8.314*573x} l n ( 3.75 ∗ 1 0 − 4 7.50 ∗ 1 0 − 4 ) = 8.314 ∗ 573 x 101000 ( x − 573 )
0.69 = 101000 x − 57873000 4764 x 0.69 = {101000x - 57873000 \over 4764x} 0.69 = 4764 x 101000 x − 57873000
3287.16 x = 101000 x − 57873000 3287.16x = 101000x - 57873000 3287.16 x = 101000 x − 57873000
57873000 = 97712.84 x 57873000 = 97712.84x 57873000 = 97712.84 x
x = 592.27 K = 592 K x = 592.27 K = 592K x = 592.27 K = 592 K
T = 592 K = 31 9 0 C T = 592K = 319^0C T = 592 K = 31 9 0 C
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