Question #194390

The rate of a standard reaction is 0.00543 M/s at 40 oC. What will the rate be if the temperature is doubled?


A. 0.0109 M/s

B. 0.0217 M/s

C. 0.0434 M/s

D. 0.0869 M/s

E. All of the Above


1
Expert's answer
2021-05-18T06:04:03-0400

R1=0.00543T1=40+273=313  KT2=80+273=353  KR2R1=Q10T2T110R_1 = 0.00543 \\ T_1 = 40 + 273 = 313 \;K \\ T_2 = 80 +273 = 353 \;K \\ \frac{R_2}{R_1} = Q_{10}^{ \frac{T_2-T_1}{10}}

For most biological systems Q10 is from 2 to 3.

Suppose Q10=2Q_{10} = 2

R20.00543=235331310R20.00543=24R20.00543=16R2=0.00543×16=0.0869\frac{R_2}{0.00543} = 2^{ \frac{353-313}{10}}\\ \frac{R_2}{0.00543} = 2^{ 4} \\ \frac{R_2}{0.00543} = 16 \\ R_2 = 0.00543 \times 16 = 0.0869

Answer: D. 0.0869 M/s


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Comments

Assignment Expert
19.05.21, 09:36

Dear Lara,

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lara
19.05.21, 04:00

thank you so much

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