Calculate the equilibrium constant temperature given. (a) I 2 ( s ) + Cl 2 ( g ) → 2 ICl ( g ) ( T = 100 ° C ) (b) H 2 ( g ) + I 2 ( s ) → 2 HI ( g ) ( T = 0.0 ° C ) (c) 2CS 2 ( g ) + 3 Cl 2 ( g ) → CCl 4 ( g ) + S 2 Cl 2 ( g ) ( T = 125 ° C ) (d) 2SO 2 ( g ) + O 2 ( g ) → 2 SO 3 ( g ) ( T = 675 ° C ) (e) CS 2 ( g ) → CS 2 ( l ) ( T = 90 ° C )
Calculate the equilibrium constant temperature given. (a) I 2 ( s ) + Cl 2 ( g ) → 2 ICl ( g ) ( T = 100 ° C ) (b) H 2 ( g ) + I 2 ( s ) → 2 HI ( g ) ( T = 0.0 ° C ) (c) 2CS 2 ( g ) + 3 Cl 2 ( g ) → CCl 4 ( g ) + S 2 Cl 2 ( g ) ( T = 125 ° C ) (d) 2SO 2 ( g ) + O 2 ( g ) → 2 SO 3 ( g ) ( T = 675 ° C ) (e) CS 2 ( g ) → CS 2 ( l ) ( T = 90 ° C )
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What I Have Learned
Directions: Given the following reaction and the stress applied in each
reaction, answer the question below.
A. H2(g) + Cl2(g) 2 HCl(g)
Stress applied: Decreasing the pressure
1. What is the Keq expression?
2. What will be the effect in the number of moles of HCl(g)?
3. What will be the Equilibrium Shift or the reaction?
B.
Fe3O4(s) + 4 H2(g) + heat 53 Fe(s) + 4 H₂O(g)
Stress applied: Increasing the temperature
1. What is the Keq expression?.
2. What will be the effect in the volume of water vapor collected?
3. What will be the Equilibrium Shift or the reaction?
C. 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) + heat
Stress applied: Increasing the volume of the container
1. What is the Keq expression?.
2. What will be the effect in the amount of H₂O?
3. What will be the Equilibrium Shift or the reaction?
Consider the solubility products (Ksp values) for the following compounds:SrSO4 (Ksp = 7.6 x 10−7), BaSO4 (Ksp = 1.5 x 10−9), SrCO3 (Ksp = 7.0 x 10−10), BaCO3 (Ksp = 1.6 x 10−9)Which anion is the harder base, CO32− or SO42−? Justify your answer.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
The Laws of Thermodynamics, Entropy, and Gibbs Free Energy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8N1BxHgsoOw;License: Standard YouTube License, CC-BY