Question

Cryosurgical techniques involve lowering the temperature of the patient's body before proceeding with the procedure. Knowing that the activation energy of the myocardial beats is of the order of 30 kJ, estimate the number of beats per minute at 22.2 ° C knowing that the number of beats per minute at 37°C is 75 beats per minute.

Answer 1

Step 1 - Analysis of the information

Since our heart beat must be consequence of a biochemical reaction, we can take the number of beats per minute as a proxy of the velocity of the reaction. Consequently, since we expect the concentration of all the chemicals to remain constant, the beats per minute can be taken as a proxy for the reaction velocity constant (k).

Step 2 - Formulate the problem in terms of the relation of the velocity constant to temperature

We know that the velocity constant relates to the temperature according to the following formula:

`k=Ae^{(-E_a)/(RT)}`

We can, then, for each beats per minute, assume that there's a constant k, which relates to the temperature in each case:

`\begin{gathered} k_1=Ae^{(-Ea)/(RT_(_1))} \\ k_2=Ae^{(-Ea)/(RT2)} \end{gathered}`

Step 3 - Proceed to the math

We can see in the equations above the factor A, which we do not know. That is not a problem, since we can eliminate it by dividing both equations. We have the following relation:

`(k_1)/(k_2)=\frac{Ae^{(-Ea)/(RT_(_1))}}{Ae^{(-Ea)/(RT_(_2))}}=\frac{e^{(-Ea)/(RT_(_1))}}{e^{(-Ea)/(RT_(_2))}}`

This equation can be further simplified, by aplying the rule of powers of the same base:

`(k_1)/(k_2)=e\lbrack(E_a)/(R)((1)/(T_2)-(1)/(T_1))\rbrack`

We can, now, eliminate the exponential part of the equation by aplying log (ln) in both sides:

`\ln (k_1)/(k_2)=(E_a)/(R)((1)/(T_2)-(1)/(T_1))`

Finally, as we saw in step 1, the concentration being the same, the velocity will be proportional to the velocity constant. We can, therefore, substitute the velocity constant for the velocity of each reaction, i.e., beats per minute:

`\ln (v_1)/(v_2)=(E_a)/(R)((1)/(T_2)-(1)/(T_1))`

Step 4 - Substitute the values in the equation

From the exercise, we know that v1 = 75 bpm, T1 = 37°C (310K), T2 = 22.2°C (295K), Ea = 30000 J and R = 8.314 J/K.mol. Let's substitute these values and work with the equation:

`\ln \frac{75_{}}{v_2}=\frac{30000_{}}{8.314}(\frac{1}{295_{}}-\frac{1}{310_{}})`
`\ln \frac{75_{}}{v_2}=3608*1.6*10^(-4)\rightarrow\text{ }\ln \frac{75_{}}{v_2}=0.57`

Now let's apply the definition of log, so we have:

`\ln \frac{75_{}}{v_2}=0.57\text{ }\rightarrow\text{ }(75)/(v_2)=e^(0.57)`

Finally, we can obtain the velocity v2 in bpm:

`v_2=\frac{75^{}}{1.76}\approx42\text{ beats per minute}`

So, at 22.2°C, the heart would beat at 42 beats per minute.