Question

Bottom chart is correct, but would like an explanation for how to find the answers to E and F. Please show work. Rate law is k=2.0x10^-2 L/(molxhr)

C Wha percent of the CHCHO remains after 200 hours C? D. What is the half-life of CH CHO iro.10 mole is injected into a 10-L vessel at 527 E How many moles of CH,CHO remain after 200 hours? F. What percent of the CHaCHO remains after 200 hours? Complete the following table: Initial moles | [CH3CHOJo | [CH3CHO] | Moles of | % CH3CHO CH3CHO after 200 hrCH3CHO Remaining after 200 hr | 0.071 M | 0.0i l mol | 7 1% |o.coqOM| 0.0aGM01 Condition1 0.10 0.10 Condition 2 0.10 0.010 -0% What conclusion can you draw about the initial concentration of a reactant in a second order reaction?

Answer 1

Ans. #E.

#F. Condition 1:     

% Remaining [CH₃CHO] = (Remaining [CH₃CHO] / Initial [CH₃CHO]) x 100

                                    = (0.714 M / 0.10 M) x 100

                                    = 71.4 %

# Condition 2:         

% Remaining [CH₃CHO] = (Remaining [CH₃CHO] / Initial [CH₃CHO]) x 100

                                    = (0.0096 M / 0.010 M) x 100

                                    = 96 %

# Conclusion: From the available data, greater is the initial concetration of reactant, smaller would be the remaining concetration (%) of reactant after specified time.