$(30$ points$)$ Reconsider the reaction equilibrium problem from $"4-$ Root$-$Finding Examples $2"$:

$A+2$BharrC$+D$

$K_C=\frac{([C_0]+)([D_0]+)}{([A_0]-)([B_0]-2{)}^2}$

$K_C(T)=15$exp$(\frac{1\mathrm{.}5{10}^6}{8\mathrm{.}314}[\frac{1}{310\mathrm{.}15}-\frac{1}{T[K]}])$

a$.(15$ points$)$ There are several ways to convert this problem into $f()=0.$ For consistency, let's rearrange problem into:

$K_C(T)([A_0]-)([B_0]-2{)}^2-([C_0]+)([D_0]+)=f()=0$

Note, $K_C(T)$ is defined by the function of temperature above. This is NOT $K_C$ times $T.$

Which method is faster for finding the root at $T=310\mathrm{.}15K,$ bisection or false position? Why? Hint: Look at a plot of $f()$ versus $.$ We know from the example that the root is approximately $=0\mathrm{.}4271M.$

b$.(15$ points$)$ We wish to drive this reaction $"$to the right". That is$,$ we wish to increase our concentration of products. Assuming the feed is $1\mathrm{.}0M$ in both A and $B$ and there is no product initially $($same conditions as the example problems$),$ what temperature gives a final concentration of $[C]=[D]=0\mathrm{.}499M?$

$3.(30$ points$)$ Reconsider the distillation example from $"4-$Root$-$Finding Examples $2".$ Generate a plot of $T$ versus $\frac{V}{F}.\frac{V}{F}$ should range from $0$ to $1$ with at least $20$ points.