The equilibrium constant of the reaction is 282. Option D
<h3>What is equilibrium constant?</h3>
The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.
Concentration of bromine = 0.600 mol /1.000-L = 0.600 M
Concentration of iodine = 1.600 mol/1.000-L = 1.600M
In this case, we must set up the ICE table as shown;
Br2(g) + I2(g) ↔ 2IBr(g)
I 0.6 1.6 0
C -x -x +2x
E 0.6 - x 1.6 - x 1.190
If 2x = 1.190
x = 1.190/2
x = 0.595
The concentrations at equilibrium are;
[Br2] = 0.6 - 0.595 = 0.005
[I2] = 1.6 - 0.595 = 1.005
Hence;
Kc = [IBr]^2/[Br2] [I2]
Kc = ( 1.190)^2/(0.005) (1.005)
Kc = 282
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Answer:
510 g NO₂
General Formulas and Concepts:
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
- Reading the Periodic Table
- Writing Compounds
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)
<u>Step 2: Define conversions</u>
Avogadro's Number
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol
<u>Step 3: Use Dimensional Analysis</u>
<u />
= 511.901 g NO₂
<u>Step 4: Check</u>
<em>We are given 2 sig figs. Follow sig fig rules.</em>
511.901 g NO₂ ≈ 510 g NO₂
Answer:
the answer has been given below have a good day
Explanation:
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