Experimental Determination of an Equilibrium Constant The decomposition of HI(g) is represented by the equation 2HI(g) = H₂(g) + 1₂(g) The following experiment was devised to determine the equilibrium constant of the reaction. HI(g) is introduced into five identical 400-cm³ glass bulbs, and the five bulbs are maintained at 623 K.The amount of 1₂ produced over time is measured by opening each bulb and titrating the contents with 0.0150 M Na2S₂O3(aq). The reaction of I₂ with the titrant is I2 + 2Na2S₂O3 →→ Na₂S4O6 + 2Nal Experimental data Data for the experiment are provided in this table. Initial mass of HI Time Volume of titrant (hours) (g) (mL) Bulb 1 2 3 4 5 0.300 0.320 0.315 0.406 0.280 bulb 1 bulb 2 bulb 3 Part A In which bulb would you expect the composition of gases to be closest to equilibrium? Ⓒbulb 4 bulb 5 2 4 12 20 40 Submit 20.96 27.90 32.31 41.50 28.68 My Answers Give Up

Experimental Determination of an Equilibrium Constant The decomposition of HI(g) is represented by the equation 2HI(g) = H₂(g) + 1₂(g) The following experiment was devised to determine the equilibrium constant of the reaction. HI(g) is introduced into five identical 400-cm³ glass bulbs, and the five bulbs are maintained at 623 K.The amount of 1₂ produced over time is measured by opening each bulb and titrating the contents with 0.0150 M Na2S₂O3(aq). The reaction of I₂ with the titrant is I2 + 2Na2S₂O3 →→ Na₂S4O6 + 2Nal Experimental data Data for the experiment are provided in this table. Initial mass of HI Time Volume of titrant (hours) (g) (mL) Bulb 1 2 3 4 5 0.300 0.320 0.315 0.406 0.280 bulb 1 bulb 2 bulb 3 Part A In which bulb would you expect the composition of gases to be closest to equilibrium? Ⓒbulb 4 bulb 5 2 4 12 20 40 Submit 20.96 27.90 32.31 41.50 28.68 My Answers Give Up

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter12: Chemical Equilibrium

Section: Chapter Questions

Problem 12.36PAE: The experiment in Exercise 12.33 was redesigned so that the reaction started with 0.15 mol each of...

See similar textbooks

Similar questions

The experiment in Exercise 12.33 was redesigned so that the reaction started with 0.15 mol each of N2 and O2 being injected into a 1.0-L container at 2500 K. The equilibrium constant at 2500 K is 3.6 X 10“’. What was the composition of the reaction mixture after equilibrium was attained? The following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) *2 2 NO K = 4.1 X IO-4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L-’ ofO2, what were the equilibrium concentrations of all species?
Again the experiment in Exercise 12.33 was redesigned. This time, 0.15 mol each of N, and O2 was injected into a 5.0-L container at 2500 K, at which the equilibrium constant is 3.6 X 10-?. What was the composition of the reaction mixture at equilibrium? l'he following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) *2 2 NO K = 4.1 X IO-4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L-’ ofO2, what were the equilibrium concentrations of all species?
Consider 0.200 mol phosphorus pentachloride sealed in a 2.0-L container at 620 K. The equilibrium constant, Kc, is 0.60 for PCl5(g) PCl3(g) + Cl2(g) Calculate the concentrations of all species after equilibrium has been reached.
A common type of reaction we will study is that having a very small K value (K 1). Solving for equilibrium concentrations in an equilibrium problem usually requires many mathematical operations to be perfomed. However, the math involved when solving equilibrium problems for reactions having small K values (K 1) is simplified. What assumption is made when solving the equilibrium concentrations for reactions with small K values? Whenever assumptions are made, they must be checked for validity. In general, the 5% rule is used to check the validity of assuming x (or 2 x, 3x, and so on) is very small compared to some number. When x (or 2 x. 3x. and so on) is less than 5% of the number the assumption was made against, then the assumption is said to be valid. If the 5% rule fails, what do you do to solve for the equilibrium concentrations?
The following equilibrium was studied by analyzing the equilibrium mixture for the amount of H2S produced. Sb2S3(s)+3H2(g)2Sb(s)+3H2S(g) A vessel whose volume was 2.50 L was filled with 0.0100 mol of antimony(III) sulfide, Sb2S3, and 0.0100 mol H2. After the mixture came to equilibrium in the closed vessel at 440C, the gaseous mixture was removed, and the hydrogen sulfide was dissolved in water. Sufficient lead(II) ion was added to react completely with the H2S to precipitate lead(II) sulfide, PbS. If 1.029 g PbS was obtained, what is the value of Kc at 440C?
Describe a nonchemical system that is in equilibrium, and explain how the principles of equilibrium apply to the system.
In the reaction in Exercise 12.33, another trial was carried out. The reaction began with an initial concentration of N2 equal to the initial concentration of NO. Each had a concentration of 0.100 mol L-1. WTat were the equilibrium concentrations of all species? The following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) ^2 NO K = 4.1 X 10~4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L'1 ofO2, what were the equilibrium concentrations of all species?
The density of trifluoroacetic acid vapor was determined at 118.1 C and 468.5 torr, and found to be 2.784 g/L. Calculate KC for the association of the acid.
The diagram represents an equilibrium mixture for the reaction N2(g) + O2(g) ⇌ 2 NO(g) Estimate the equilibrium constant.