An interaction between two subunits of a protein was determined to have a delta G of -57.05 kj/mol. What is the Keq for the reaction at 25 degrees celsius?

Apr 8, 2016

An assumption is made here that either the physiological temperature around the protein is ${25}^{\circ} \text{C}$, or the protein was isolated outside of physiological conditions and placed into pseudo-physiological conditions at ${25}^{\circ} \text{C}$.

So this may not be a realistic assumption, but we'll go with it so you get the "right" answer in accordance with your textbook.

Suppose $\Delta {G}^{\circ} = - \text{57.05 kJ/mol}$ and $\text{T = 298.15 K}$ (room temperature).

Then, we have the following equation to consider:

$\setminus m a t h b f \left(\Delta G = \Delta {G}^{\circ} + R T \ln Q\right)$

where:

• $\Delta G$ is the Gibbs' free energy and $\Delta {G}^{\circ}$ is the same thing, but at ${25}^{\circ} \text{C}$ and $\text{1 bar}$. Note that in nonstandard conditions, $\Delta {G}^{\circ}$ describes exergonicity or endergonicity, not equilibrium or spontaneity.
• $R$ is the universal gas constant, $8.314472 \times {10}^{- 3} \text{kJ/mol"cdot"K}$.
• $T$ is the temperature in $\text{K}$.
• $Q$ is the reaction quotient, i.e. the "not-yet-equilibrium" constant.

When we are at equilibrium and $\text{298.15 K}$ like the problem asks, $\Delta G = 0$, and $Q = K$. Thus, we get:

$\textcolor{g r e e n}{\Delta {G}^{\circ} = - R T \ln {K}_{e q}}$

So, to calculate ${K}_{e q}$, we divide $- R T$ over and exponentiate.

$\textcolor{b l u e}{{K}_{e q}} = {e}^{- \Delta {G}^{\circ} \text{/RT}}$

= e^(-(-"57.05 kJ"cdot"mol"^(-1)"/"8.314472xx10^(-3) "kJ"cdot"mol"^(-1)"K"*"298.15 K")

$= {e}^{23.01}$

$\approx \textcolor{b l u e}{9.879 \times {10}^{9}}$