A reversible chemical reaction

$2$A$+$B C

can be characterized by the equilibrium relationship

K $=($C$\_$C$)/[(($C$\_$A$)^2)*($C$\_$B$)$

where the nomenclature C$\_$i represents the concentration of constituent i$.$ Suppose that we define a variable x as representing the number of moles of C that are produced. Conservation of mass can be used to reformulate the equilibrium relationship as

K $=($C$\_$C$,0+$ x$)/[($C$\_$A$,0-2$x$)^2($C$\_$B$,0-$ X$)]$

where the subscript $0$ designates the initial concentration of each constituent. If K$=0\mathrm{.}015,$ C$\_$A$,0=42,$ C$\_$B$,0=29,$ C$\_$C$,0=7.$ Determine the value of x$.$ Use the bisection method to solve for the root, with initial guesses of X$\_$L $=0$ and X$\_$U $=20.$ Starting with the initial values, go through FOUR A reversible chemical reaction,

$2A+BC$

can be characterized by the equilibrium relationship

$K=\frac{c_C}{C_A^2{c}_B}$

$K=\frac{(c_{C,0}+x)}{(c_{A,0}-2x{)}^2(c_{B,0}-x)}$

Answer:

$x_{L,0}=$

$x_{U,0}=$

$x_{R,0}=$

$x_{L,1}=$

$x_{U,1}$

$x_{R,1}=$

$x_{L,2}=$

$x_{U,2}=$

$x_{R,2}$

$x_{L,3}$

$x_{U,3}=$

$x_{R,3}=$

$x_{L,4}=$

$x_{U,4}=$

$x_{R,4}=$iterations of the bisection method, and input your lower$/$upper boundaries and root estimate for each iteration.