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initial [I2] = number of mol / volume
= 0.0461 mol / 2.27 L
= 0.0203 M
I2 (g) <---> 2I (g)
0.0203
0 (initial)
0.0203-x
2x (at equilibrium)
Kc= [I]^2 / [I2]
3.80*10^-5 = (2x)^2 / (0.0203-x)
since Kc is small, x can be ignored as compared to 0.0203
Above expression thus becomes,
3.80*10^-5 = (2x)^2 / (0.0203)
(2x)^2 = 7.714*10^-7
x = 4.39*10^-4 M
[I2] = 0.0203-x = 0.0203 - 4.39*10^-4 = 0.0199 M
[I] = 2x= 2*4.39*10^-4 = 8.78*10^-4 M
21.Be sure to answer all parts. The dissociation of molecular iodine into iodine atoms is represented...
The dissociation of molecular iodine into iodine atoms is represented as I2(g) ⇌ 2I(g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 × 10−5. Suppose you start with 0.0458 mol of I2 in a 2.32−L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of I2? __M What is the equilibrium concentration of I? __M
Please help me with this question and help me understand
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Be sure to answer all parts. The dissociation of molecular iodine into iodine atoms is represented as I2(g) - 21g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 × 10-5. Suppose you start with 0.0458 mol of I2 in a 2.33-L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of I2? What is the equilibrium...
3 attempts left Check my work Be sure to answer all parts. The dissociation of molecular iodine into iodine atoms is represented as 126) = 2(g) At 1000 K, the equilibrium constant K for the reaction is 3.80 x 10 . Suppose you start with 0.0451 mol of I, in a 2.33-L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of IZ? M What is the equilibrium concentration of I?...
Be sure to answer all parts. The dissociation of molecular lodir Is represented as 13C) = 2(g) brium constant for the reaction is 3.80 X 10. Suppose you start with 0.0454 mol of 1 in a 2.33-L flask the concentrations of the gases at equilibrium? Wha the equilibrium concentration of 1,? What is the equilibrium concentration of I?
the equilibrium constant for the dissociation of iodine molecules to iodine atoms I2 (g) arrow 2I(g) is 3.76x10^-3 at 1000 K. suppose 0.170 mol of I2 is placed in a 18.7 L flask at 1000 K. what are the concentrations of I2 and I when the system comes to equilibrium?
1)Consider the following equilibrium at 972 K for the dissociation of molecular iodine into atoms of iodine. I2(g) ----> 2 I(g); Kc = 1.40 ? 10?3 Suppose this reaction is initiated in a 3.4 L container with 0.067 mol I2 at 972 K. Calculate the concentrations of I2 and I at equilibrium. * I got I2= 0.0135M and I= 0.0124 is this right?* 2)Consider the following equilibrium. NH3(aq) + H2O(l) -----> NH4+(aq) + OH ?(aq) What will happen to the...
9. The equilibrium constant(K.) for the gaseous dissociation of Iodine molecules to lodine atoms (shown below) 12(8) 21(8) is 3.76 x 10-3 at 1000K. If 0.15 mole of 12(g) is placed in a 12.3-L flask at 1000 K, what are the concentrations of 12(g) and I(g) when the reaction comes to equilibrium?
Be sure to answer all parts. Kc for the reaction of hydrogen and iodine to produce hydrogen iodide. H2(g) + I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Calculate the equilibrium concentrations of H2, I2, and HI at 430°C if the initial concentrations are [H2] = [I2] = 0 M, and [HI] = 0.419 M. [H2] = [I2] = [HI] =
Be sure to answer all parts. Kc for the reaction of hydrogen and iodine to produce hydrogen iodide. H2(g) + I2(g) ⇌ 2HI(g) is 54.3 at 430°C. Calculate the equilibrium concentrations of H2, I2, and HI at 430°C if the initial concentrations are [H2] = [I2] = 0 M, and [HI] = 0.349 M. [H2] = M [I2] = M [HI] = M
Molecular iodine, I2 (g), dissociates into iodine
atoms at 652 K, with a first order rate constant of 0.452
s-1.
a. What is the half life for this reaction?
[ Select ]
["1.53 s", "0.31 s", "0.079 s", "0.39 s"]
b. If you start with 0.35 M I2 at this temperature,
how long will it take for the concentration of I2 to
reach...