University:
Carleton College
Course:
CHEM 122 | Introduction to Chemistry
Academic year:

2022
Views:

88

Pages:

2
Author:

StellarSymphony

2 C4H6(g)  C8H12(g) At high temperatures the compound C4H6 (1,3-butadiene) reacts according to the equation above. The rate of the reaction was studied at 625 K in a rigid reaction vessel. Two different trials, each with a different starting concentration, were carried out. The data were plotted in three different ways, as shown below. (a) For trial 1, calculate the initial pressure, in atm, in the vessel at 625 K. Assume that initially all the gas present in the vessel is C4H6 . For trial 1, n = 0.020 mol/L (or assume the volume of the vessel is 1.0 L; V the number of moles of C4H6 in the vessel would then be 0.020 mol). PV = nRT (0.020 mol)(0.08206 L atm mol -1 K -1 )(625 K) nRT = 1.0 atm P = = V 1.0 L 1 point is earned for a correct setup. 1 point is earned for the correct answer. (b) Use the data plotted in the graphs to determine the order of the reaction with respect to C4H6 . Second order (because the plot of 1/[C4H6] is a straight line). 1 point is earned for the correct order. (c) The initial rate of the reaction in trial 1 is 0.0010 mol/(Ls). Calculate the rate constant, k, for the reaction at 625 K. From the second-order rate law (differential form): rate = k[C4H6]2  k = 0.0010 mol/(L× s) rate = = 2.5 L/(mol× s) 2 ([C4 H 6 ]) (0.020 mol/L)2 OR From the second-order rate law (integrated form): 1 1 = 2kt + [C4 H 6 ]0 [C4 H 6 ] t The coefficient of t is equal to 2k because of the reaction stoichiometry. 1 The slope of the line in the plot of versus time is 2k. [C4 H 6 ] Thus slope = 5.0 L/(mol·s) = 2k , therefore k = 2.5 L/(mol·s). Note: Students who choose the second method of determining k but omit the factor of 2, thereby getting an answer of 5.0 L/(mol·s), still earn the point. 1 point is earned for the correct value.