-Find the appropiate equilibrium constant (Kp) for the reaction below under the
ID: 800631 • Letter: #
Question
-Find the appropiate equilibrium constant (Kp) for the reaction below under the following conditions.
1/2SnO2 (s) + H2(g) <--> H2O(g) + 1/2 Sn(s)
a)At 650 degrees celsius the equilibrium steam - H2 mixture was 38 % H2 by volume.
b)At 800 degrees celsius the equilibrium steam- H2 mixture was 22% H2 by volume.
c)Would you run this reaction at high or low temperature for the most efficient reduction of Tin (IV)
-At 35degrees celsius, Kc=1.6*10^-5 for the reaction
2NOCl(g) <--> 2NO(g) +Cl2(g)
Caluculate the concentrations of all species at equilibrium for each of the following original mixtures
a) 1.0 mole NOCl in a 1.0 L container
b) 1.00 mol NO and 0.50 mol Cl2 in a 1.0 L container
c)1.00 mol NO and 0.75 mol Cl2 in a 1.0 L container
-Use the following reaction for the next two problems:
2A (g)+B(g) <--> 3C (g) + D(g)
*Determine Kc if 6 .0 moles of A added to 6.0 moles of B in a 2.0 L flask yields 0.10 moles of D at equilibrium.
*Determine the equilibrium concentration of D from the following initial concentrations.
a) 6.0 moles A +6.0 moles B in 1.0L flask.
b)3.0 mole A +3.0 moles B in a 2.0L flask.
c)6.0 moles C + 6.0 moles D in a 2.0L flask.
d) 2.0 moles A + 1.0 moles B + 3.0 moles C+ 1.0 moles D in a 2.0L flask.
-A solution is prepared having having the following initial concentrations:
[Fe3+]=[Hg2 2+] = 0.5000M; [Fe3+]= [Hg2+]=0.03000M
*The following reaction occurs among the ions ata certain temperature.
2Fe3+ (aq) +Hg2 2+ (aq)<-->2Fe2+(aq) +2Hg2+(aq) Kc=9.14*10^-8
What will be the ion concentrations when equilibrium is established?
Explanation / Answer
2)The K refers to the equlibrium concentrations of the 3 gases as follows
[NO]^2[Cl2]/[NOCl]^2 = 1.6x10-5
Equilibrum can be attained from either the reactants end (no products to start), the products end (no reactant to start), or 2 of the 3 gases. Your problems illustrate this. I'll show you (d), hopefully you can work out the others.
When you have all 3 gases, it helps to guess which way the reaction is going. If you "plug" in the initial reactants, the result is 1.9, which is much higher than K. So, suspect that the reaction will run backwards. So let x be the moles/L of chlorine lost at equilibrium, so at equilibrium, we have 1.9-x moles. Since all reactions are in a liter flask, thats where we get the per liter. Then the other concentrations at equilibrium are
1.9-2x for NO and 1.9+2x for NOCl.
Substituting, (1.9-x)(1.9-2x)^2 /(1.9+2x)^2 =K
You can solve this by algebra or by trial and error. For example, if x = 0.9, we have (1)(0.01)/3.6^2.
which is about 7x10-4. From the term for NO, x cant exceed 0.95, otherwise the expression goes to zero. So, the answer should be pretty close to 0.95 for Cl2, 3.7 for NOCl, and somewhere around 0.04 mole/L.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.