2NoBr>2NO+Br2 Kc=0.064
At equilibrium a 1.00L flask contains 0.030mole NOBr and 0.030mole of NO.How many moles of Br2 are present.
Kc = (NO)^2 (Br2)/(NOBr)^2
Substitute and solve for the one unknown.
To determine the number of moles of Br2 present at equilibrium, we need to use the equilibrium constant (Kc) and the given initial amounts of substances.
The balanced chemical equation for the reaction is:
2NOBr ⇌ 2NO + Br2
The equilibrium expression is given as:
Kc = [NO]^2 * [Br2] / [NOBr]^2
Given:
Initial moles of NOBr = 0.030 moles
Initial moles of NO = 0.030 moles
Total volume of the flask = 1.00 L
Since the initial moles of NOBr and NO are equal, we can assume that the reaction starts with 0.030 moles of NOBr and NO, and 0 moles of Br2.
Let x be the number of moles of Br2 at equilibrium.
At equilibrium, the moles of NOBr will decrease by 2x, and the moles of NO and Br2 will increase by x.
The expression for the equilibrium constant can now be written as:
Kc = ([NO] + x)^2 * (x) / ([NOBr] - 2x)^2
Given that Kc = 0.064, we can substitute the values into the equation and solve for x.
0.064 = (0.030 + x)^2 * (x) / (0.030 - 2x)^2
Simplifying the equation, we have:
0.064 = (0.030 + x)^2 * (x) / (0.030 - 2x)^2
0.064 * (0.030 - 2x)^2 = (0.030 + x)^2 * (x)
0.00192 - 0.128x + 4x^2 = 0.0009 + 0.06x + x^2
3x^2 - 0.188x + 0.00102 = 0
This quadratic equation can be solved using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
where a = 3, b = -0.188, and c = 0.00102.
Solving the equation, we find two possible values for x: x = 0.025 moles or x = 0.0034 moles.
Since the moles of Br2 cannot be negative, we discard the negative value.
Therefore, at equilibrium, the number of moles of Br2 present in the 1.00 L flask is approximately 0.0034 moles.
To determine the number of moles of Br2 present at equilibrium, we need to use the equilibrium constant (Kc) expression and the given concentrations of NOBr and NO.
The balanced equation for the reaction is:
2NOBr -> 2NO + Br2
The equilibrium constant expression is given by:
Kc = [NO]^2 * [Br2] / [NOBr]^2
Let's solve for [Br2]:
Kc = [NO]^2 * [Br2] / [NOBr]^2
Given:
[NOBr] = 0.030 mole
[NO] = 0.030 mole
Kc = 0.064
Substituting these values into the equation, we get:
0.064 = (0.030)^2 * [Br2] / (0.030)^2
Simplifying the equation, we have:
0.064 = [Br2]
Therefore, the number of moles of Br2 present at equilibrium is 0.064 moles (or 0.064 mol).