# Calculating enthalpy per gram mole of oxygen bond in haemoglobin

A solution of $\pu{5 g}$ of haemoglobin in $\pu{100 cm^3}$ of solution shows a temperature raise of $\pu{0.031 ^\circ C}$ for complete oxygenation. Each mole of haemoglobin binds $\pu{4 mol}$ of oxygen. If the heat capacity of the solution is $\pu{4.18 J K^{-1} cm^{-3}}$, calculate $\Delta H$ per g of oxygen bond.

Attempt:

• for $\pu{100 cc}$, heat capacity is $\pu{418 J K^{-1}}$
• for $\pu{\frac{5}{64000}mol}$ of haemoglobin, $C\Delta T=\pu{12.952 J}$ is released
• hence for $\pu{1 mol}$, $\pu{165862.4 J}$ energy is released
• This energy is released when four bonds are formed with oxygen. Hence for 1 bond, $\pu{41.47 kJ}$ energy is released.

The answer given in the book is $\pu{-41.47 kJ}$.

The heat flows from each Hb molecule during the reaction at constant p and T, so $q_{p, rxn} = \Delta H_{rxn}\cdot\Delta X$.

The energy of the reaction is $\Delta H_{rxn}$ as per mole of reaction.

The reaction itself is Hb + 4O$_{2}\rightarrow$ Hb$\cdot$(O$_{2}$)$_{4}$.

The reaction is exothermic.

The number of moles of Hb present is $$n_{\rm Hb} = {5\over 64,000} = 0.078\;{\rm mmol},$$

all of which reacts: $\Delta X = -\Delta\,n_{\rm Hb} = n_{\rm Hb}$.

That much heat raises the temperature of the water (assuming all the heat is contained in the water):

$$0 = \Delta H_{\rm water} + q_{p, rxn} = C_{p}\Delta T + q_{p, rxn}$$ from which it follows that $$q_{p, rxn} = -C_{p}\Delta T = -100\times 4.18\times 0.031 = -13\;{\rm J}$$ and $$\Delta H_{rxn} = {q_{p}\over\Delta X} = {-13\;{\rm J}\over 0.078\;{\rm mmol}}= -167\;{{\rm kJ}\over{\rm mol-rxn}}$$ There are 4 moles of O$_{2}$ used so $$\nu_{\rm O_{2}} = 4\;{{\rm O_{2}}\over{\rm rxn}}$$ and on an oxygen basis $$\Delta H_{rxn} = -42\;{\rm kJ}$$