Question

The rate of a standard reaction is 0.01840 M/s at 25 oC. It is determined that this is too fast, and that the rate should be reduced to 0.0046 M/s. What temperature should the reaction be run at to achieve this?

A. 45 oC
B. 20 oC
C. 15 oC
D. 5 oC
E. 0 oC

Answer 1

Question #79449

The rate of a standard reaction is 0.01840 M/s at 25 oC. It is determined that this is too fast, and that the rate should be reduced to 0.0046 M/s. What temperature should the reaction be run at to achieve this?

A. 45 oC

B. 20 oC

C. 15 oC

D. 5 oC

E. 0 oC

Answer:

The right answer is D. 5 °C.

According to the equation [1]:

$$\frac {R _ {2}}{R _ {1}} = Q _ {1 0} ^ {\frac {T _ {2} - T _ {1}}{1 0}},$$

where $R_{1}$ – is the rate of reaction at $25^{\circ}C$, $R_{2}$ – is the rate of reduced reaction, $T_{1}$ – is the temperature of standard reaction ($T_{1} = 25 + 273 = 298~K$), $T_{2}$ – is the temperature of reduced reaction (In K), $Q_{10}$ – is the $Q_{10}$ temperature coefficient.

For most biological systems, the $Q_{10}$ value is ~ 2 to 3.

$$\frac {0 . 0 0 4 6}{0 . 0 1 8 4} = Q _ {1 0} ^ {\frac {T _ {2} - 2 9 8}{1 0}}$$ $$0. 2 5 = Q _ {1 0} ^ {\frac {T _ {2} - 2 9 8}{1 0}}$$

If we suggest that $Q_{10}$ is equal to 2, we get the following:

$$0. 2 5 = 2 ^ {\frac {T _ {2} - 2 9 8}{1 0}}$$

$T_{2}$ should be equal to 278, to be the solution of the equation.

$$t _ {2} = 2 7 8 - 2 7 3 = 5 ^ {\circ} \mathrm {C}.$$

So, the right answer is D. 5 °C.

Reference:

[1] https://en.wikipedia.org/wiki/Q10 (temperature_coefficient)

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