Question

When heated, the DNA double helix separates into two random-coil single strands. When cooled, the random coils reform the double helix: double helix 2 random coils.

(a) What is the sign of$\mathbf{∆}\mathbf{\text{s}}$for the forward process? Why?

(b) Energy must be added to overcome H bonds and dispersion forces between the strands. What is the sign of$\mathbf{∆}\mathit{G}$for the forward process when$\mathbf{T\Delta S}$is smaller thanrole="math" localid="1663303287852" $\mathbf{∆}\mathbf{\text{H}}$?

(c) Write an expression that shows T in terms of$\mathbf{∆}\mathbf{\text{H}}$and$\mathbf{∆}\mathbf{\text{s}}$when the reaction is at equilibrium. (This temperature is called the melting temperature of the nucleic acid.)

Short answer

a) For the forward process, set $\mathrm{\Delta S}>0$.

b) For the forward process, Gibb's energy is positive.

c) Expression of melting temperature$\mathrm{T}=\frac{\mathrm{\Delta H}}{\mathrm{\Delta S}}$.

Step-by-step solution

Definition of DNA

Deoxyribonucleic acid (DNA) is a polymer made up of two polynucleotide chains that coil around each other to create a double helix and carry genetic instructions for all known organisms and viruses' formation, function, growth, and reproduction.

Determining the sign of ∆s for the forward process

a) The system gets increasingly disordered as the number of 'particles' increases during the reaction to generate two random coils from a single double helix, raising the entropy of the system and making the reaction spontaneous.

As a result, for the forward process, $\mathrm{\Delta S}>0$.

Determining the sign of ∆G for the forward process

b) The $\mathrm{\Delta G}$can be written as follows:

$\mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T}\times \mathrm{\Delta S}$

If $\mathrm{T}\times \mathrm{\Delta S}<<\mathrm{\Delta H}$, then

$\begin{array}{rcl}∆H-T\times ∆s& >& 0\\ ∆G& >& 0\end{array}$

Gibb's vitality is beneficial to the onward movement.

Determining the expression of T in terms of ∆H and ∆s

c) Since $\mathrm{\Delta G}=0$while the reaction is at equilibrium,

$\mathrm{\Delta H}-\mathrm{T}\times \mathrm{\Delta S}=0$

as well as the melting temperature, which can be written as

$\mathrm{T}=\frac{\mathrm{\Delta H}}{\mathrm{\Delta S}}$