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. 0.01086 M/s
B. 0.02172 M/s
C. 0.04344 M/s
D. 0.08688 M/s
E. All of the Above

Answer 1

Question #80617

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. 0.01086 M/s

B. 0.02172 M/s

C. 0.04344 M/s

D. 0.08688 M/s

E. All of the Above

Answer:

The right answer is D. 0.08688 M/s.

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 $40^{\circ}C$, $R_{2}$ – is the rate of reduced reaction, $T_{1}$ – is the temperature of standard reaction ($T_{1} = 40 + 273 = 313 K$), $T_{2}$ – is the temperature of reduced reaction ($T_{2} = 80 + 273 = 353 K$), $Q_{10}$ – is the $Q_{10}$ temperature coefficient.

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

$$\frac {R _ {2}}{0 . 0 0 5 4 3} = Q _ {1 0} ^ {\frac {3 5 3 - 3 1 3}{1 0}}$$ $$\frac {R _ {2}}{0 . 0 0 5 4 3} = Q _ {1 0} ^ {4}$$

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

$$\frac {R _ {2}}{0 . 0 0 5 4 3} = 2 ^ {4}$$ $$R _ {2} = 0. 0 0 5 4 3 * 1 6 = 0. 0 8 6 8 8 M / s$$

So, if we double the temperature of reaction, the rate of reaction has to be equal to 0.08688 M/s (D).

Reference:

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