consider the following equilibrium
2NOCl(g)--- 2 NO(g) + Cl2(g)
Initally 1.00 mol of NO and 1.00 mol of Cl2 are placed in a 5.00L container.
Calculate the initial concentrations of NOCl, NO and Cl2
NO= 0.20 mol/L
Cl2= 0.20 mol/l
NOCl=0
At equilibrium it is found that the NOCl = 0.120 mol/L calculate the value of k eq for the reaction
using ice table
change
NOCl= 0.120 mol/L
NO= -0.120 mol/L
Cl2= -0.06 mol/L
equilibrium
NOCl= 0.120 mol/L
NO= 0.08 mol/L
Cl2 = 0.14 mol/L
k eq= 0.08^2 0.14/ 0.120^2= 6.2x10^-2
if you could check this question out and point out any errors. I appreciate your time. thanks
it looks ok to me.
Based on the information you provided, it appears that you have made some errors in your calculations. Let's go through it step by step:
To calculate the initial concentrations of NO, Cl2, and NOCl, we start with the given amounts of 1.00 mol of NO and 1.00 mol of Cl2 in a 5.00L container.
Since the volume is given in liters, the initial concentration of NO and Cl2 can be calculated by dividing the number of moles of each substance by the volume of the container:
NO initial concentration = 1.00 mol / 5.00 L = 0.20 mol/L
Cl2 initial concentration = 1.00 mol / 5.00 L = 0.20 mol/L
For NOCl, you mentioned that the initial concentration is 0, but based on the balanced equation, there should be some NOCl present initially. Since the stoichiometry of the reaction indicates that 2 moles of NOCl are consumed to produce 2 moles of NO and 1 mole of Cl2, we can calculate the initial concentration of NOCl using the initial concentrations of NO and Cl2:
NOCl initial concentration = (0.20 mol/L) / 2 = 0.10 mol/L
So, the correct initial concentrations should be:
NO initial concentration = 0.20 mol/L
Cl2 initial concentration = 0.20 mol/L
NOCl initial concentration = 0.10 mol/L
Moving on to calculating the equilibrium constant (K eq) using the given equilibrium concentrations of NOCl = 0.120 mol/L, NO = 0.08 mol/L, and Cl2 = 0.14 mol/L:
K eq is determined by the ratio of the molar concentrations of the products to the molar concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients. In this case, the balanced equation shows that the stoichiometric coefficients are 2, 2, and 1 for NOCl, NO, and Cl2, respectively.
Therefore, the correct expression for K eq using the given equilibrium concentrations is:
K eq = ((0.08 mol/L)^2 * (0.14 mol/L)) / (0.120 mol/L)^2
Evaluating this expression:
K eq = (0.0064 * 0.14) / (0.0144) = 6.16
Hence, the correct value for K eq is 6.16, not 6.2x10^-2 as you calculated.
Remember to double-check the stoichiometry and perform accurate calculations to ensure the correct answers.
Your calculations for the initial concentrations of NO, Cl2, and NOCl are correct. However, there are a few errors in your calculation of the equilibrium concentrations and the value of K eq.
When using the ICE table, the change in concentration should be subtracted from the initial concentration to obtain the equilibrium concentration. Here are the corrected calculations:
Initial concentrations:
[NOCl] = 0.00 mol/L
[NO] = 1.00 mol/L
[Cl2] = 1.00 mol/L
Change in concentration:
[NOCl] = +0.12 mol/L (not -0.12 mol/L)
[NO] = -0.12 mol/L
[Cl2] = -0.06 mol/L
Equilibrium concentrations:
[NOCl] = 0.12 mol/L
[NO] = 0.88 mol/L (1.00 - 0.12 = 0.88)
[Cl2] = 0.94 mol/L (1.00 - 0.06 = 0.94)
Now let's calculate the value of K eq using these correct equilibrium concentrations:
K eq = ([NO] ^ 2 * [Cl2]) / [NOCl]^2
K eq = (0.88^2 * 0.94) / (0.12^2) ≈ 52.0
So the correct value of K eq for the reaction is approximately 52.0, not 6.2x10^-2 as you calculated.
Please let me know if you have any further questions!