The first two pictures is for information if you need it to help you and the third picture is the problem

[HI] mol/L concentration, mol/L 5 4 3 2 3 IL 0 - 2 2H1 H2 + 12 Rate Law : k[HI] k= 6.4 . 10-9 L/(mol. s) at 500 K Initial Rate = 1.6 . 10-mol/( Ls) Rate = 2.03 . 10-'mol/(L.) [HI] = 0.56 mol/L 0 2.0 2.3 05 1.0 1.5 time, 108 Plot [HI] vs. time Plot In [HI] vs, time Plot 1/[HI] vs. time 2H1 - H+1 Start Reset 1 Characteristics of second-order reactions For a second-order reaction. [A] ---products, the rate of the reaction is given as rate = k[A], where k is the rate constant 1 and [A] is the concentration of reactant A. The integrated rate law for second-order reactions is =kt + where A) [A]. [A], is the concentration of reactant A at time t, k is the rate constant, and (A), is the initial concentration of reactant A. This equation is of the type y=mx+b. Therefore, the plot of versus time is always a straight line with a slope k and a y intercept (AJ, (A) [A], mol slope - [Alo . time, s Consider the second-order reaction: 2HI(g)+H2(g) +12(g) Use the simulation to find the initial concentration (HI), and the rate constant k for the reaction. What will be the concentration of HI after t = 4.40x1010s (HT).) for a reaction starting under the condition in the simulation? Express your answer in moles per liters to three significant figures. View Available Hint(s) ?