Q20.108CP - Chemistry: Molecular Nature Of Matter And Change

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 than  $\mathbf{∆}\mathbf{\text{H}}$? 

(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.)

Step-by-Step Solution

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 1: 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.

Step 2: 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, $\mathrm{\Delta S}>0$.

Step 3: 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.

Step 4: Determining the expression of T in terms of &#8710; H and &#8710; 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

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