Question

The rate constant of a first order reaction at
27°C is 10-3 min-1. The 'temperature
coefficient' of this reaction is 2. What is the
rate constant (in min-1) at 17 °C for this
reaction ?​

Answer 1

Answer:

rate=5×$10^{-4}$

Explanation:

for every 10° rise in temp rate doubles

here temperature decreases by 10 so divide the rate of 27° by 2

$T_{1}$=27°  $rate_{1}$=$10^{-3}$

$T_{2}$=17°   $rate_{2}$=$\frac{10^{-3} }{2}$ =5×$10^{-4}$$min^{-1}$

Answer 2

Explanation:

rate=5×10^{-4}10−4

for every 10° rise in temp rate doubles

here temperature decreases by 10 so divide the rate of 27° by 2

T_{1}T1 =27°  rate_{1}rate1 =10^{-3}10−3

T_{2}T2 =17°   rate_{2}rate2 =\frac{10^{-3} }{2}210−3 =5×10^{-4}10−4 min^{-1}

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