(d) integrate the left hand side over time between zero (start of the reaction) and time i at which the reaction has reached the equilibrium, and print out the result, (e) integrate the right hand side over x between zero (the value of x at the start of the reaction is zero) and x (the amount by which the concentrations of species a and b have changed at the equilibrium relative to a and b, respectively), and print out the result, hint: you may need to tell mathematica that x s l when integrating, (1) combine the integrated ihs and this, solve for k, and print out the result. according to the literature, the final expression for k should be krina (65) replacing k with ky+k, (equation 63) we get kz+k2 = 4 in (2) (66) (g) now solve the system of two equations, (59) and (66), to get expressions for the rate constants k, and kz in terms of t, l and x, (h) obtain the final expressions for k, and k, by replacing l in the derived expressions with equation 62). the final expressions for k, and ky show that by measuring and/or estimating the value of the equilibrium constant k for the overall reaction, and measuring the deviation x from the initial concentration for either a or b at time t, it is possible to determine the rate constants for the forward and reverse reactions given in equations (57) and (58). warning! all calculation steps must be performed in your mathematica notebook, and all items (a-d) must be clearly labelled using the print[function, or the score for this exercise will be zero! answers: (h) k, = kin(1- 3) (k+1) (x+1)+