Answer to Question #194390 in General Chemistry for lara

Question

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

Answer 1

$R_1 = 0.00543 \\ T_1 = 40 + 273 = 313 \;K \\ T_2 = 80 +273 = 353 \;K \\ \frac{R_2}{R_1} = Q_{10}^{ \frac{T_2-T_1}{10}}$

For most biological systems Q₁₀ is from 2 to 3.

Suppose $Q_{10} = 2$

$\frac{R_2}{0.00543} = 2^{ \frac{353-313}{10}}\\ \frac{R_2}{0.00543} = 2^{ 4} \\ \frac{R_2}{0.00543} = 16 \\ R_2 = 0.00543 \times 16 = 0.0869$

Answer: D. 0.0869 M/s