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
R1=0.00543T1=40+273=313 KT2=80+273=353 KR2R1=Q10T2−T110R_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}}R1=0.00543T1=40+273=313KT2=80+273=353KR1R2=Q1010T2−T1
For most biological systems Q10 is from 2 to 3.
Suppose Q10=2Q_{10} = 2Q10=2
R20.00543=2353−31310R20.00543=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 = 0.08690.00543R2=210353−3130.00543R2=240.00543R2=16R2=0.00543×16=0.0869
Answer: D. 0.0869 M/s
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