When solid NH4HS and 0.28 mol NH3(g) were placed in a 2 L vessel at 24◦C, the equilibrium NH4HS(s)⇀↽NH3(g) + H2S(g)for which Kc= 0.00016, was reached. What is the equilibrium concentration of NH3?Answer in units of mol/L
In a 2 L vessel the (NH3) initially is 0.28 mols so it is 0.28/2 = 0.14 M.
.................NH4Hs(s) ==> NH3(g) + H2S(g)
I................solid..................0.14.............0
C...............solid..................+x..............+x
E...............solid...............0.14+x.............+x
Kc = 0.00016 = (NH3)(H2S)/(NH4HS)(s)
Plug in the E line, solve for x, evaluate 0.14+x and you will obtain (NH3) in mols/L @ 24 C. Note: By definition, (NH4HS) = 1 since it is a solid. Post your work if you get stuck.
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To find the equilibrium concentration of NH3, we need to set up an ICE table and use the stoichiometry of the balanced chemical equation.
The given information tells us that we have 0.28 mol of NH3 initially present. Let's assume that x mol of NH3 reacts to form x mol of H2S.
The ICE table would be as follows:
NH4HS(s) ⇌ NH3(g) + H2S(g)
Initial: 0 0.28 0
Change: -x +x +x
Equilibrium: 0-x 0.28+x x
Using the equilibrium concentrations, we can write the expression for Kc:
Kc = ([NH3]eq * [H2S]eq) / [NH4HS]eq
Plugging in the equilibrium concentrations into the equation and substituting the given value of Kc:
0.00016 = ((0.28 + x) * x) / (0-x)
Simplifying the equation:
0.00016(0 - x) = (0.28 + x) * x
0.00016x = 0.28x + x^2
Rearranging the equation:
0 = x^2 + 0.28x - 0.00016x
0 = x^2 + 0.27984x
Using the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / 2a
x = (-0.27984 ± √((0.27984)^2 - 4(1)(0))) / 2(1)
Calculating the discriminant:
√((0.27984)^2 - 4(1)(0)) = √(0.07829) ≈ 0.27984
Using the quadratic formula:
x = (-0.27984 ± 0.27984) / 2
There are two possible values for x:
1. x = (-0.27984 + 0.27984) / 2 = 0
2. x = (-0.27984 - 0.27984) / 2 = -0.27984 / 2 ≈ -0.13992
Since we cannot have negative concentrations, we discard the second value.
Therefore, x = 0.
Substituting this value back into the equilibrium concentrations, we find:
[NH3]eq = 0.28 + x = 0.28 + 0 = 0.28 mol/L.
Therefore, the equilibrium concentration of NH3 is 0.28 mol/L.
To find the equilibrium concentration of NH3, we can use the equilibrium constant (Kc) expression and the given information.
The equilibrium constant expression for the given reaction is:
Kc = [NH3] * [H2S] / [NH4HS]
We are given that Kc = 0.00016, and we need to find the equilibrium concentration of NH3. Since NH3 is a gas, its concentration can be measured in terms of moles per liter (mol/L).
Let's define the initial number of moles of NH3 as n0(NH3). We are given that 0.28 mol of NH3(g) is placed in a 2 L vessel at equilibrium, so the initial concentration of NH3 can be calculated as:
[NH3]0 = n0(NH3) / V
= 0.28 mol / 2 L
= 0.14 mol/L
Now, let's assume that at equilibrium, the concentration of NH3 becomes [NH3] (mol/L).
Using the equilibrium constant expression, we can write the equation as:
Kc = [NH3] * [H2S] / [NH4HS]
Given that solid NH4HS is in equilibrium, we can assume its concentration remains constant. Therefore, we can write it as a constant, [NH4HS], and rewrite the equation as:
Kc = ([NH3] * [H2S]) / [NH4HS]
Rearranging the equation, we get:
[NH3] = (Kc * [NH4HS]) / [H2S]
Since we know the value of Kc (0.00016) and want to find the equilibrium concentration of NH3, we need to determine the concentration of [NH4HS] and [H2S].
Given that NH4HS is initially present as a solid, we can assume its initial concentration is zero ([NH4HS]0 = 0 mol/L). However, at equilibrium, we can assume that some amount of NH4HS dissociates to form NH3 and H2S.
Let's assume x represents the concentration of NH3 (which is the same as [NH3] at equilibrium) and y represents the concentration of H2S (which is the same as [H2S] at equilibrium).
Using the balanced equation:
NH4HS(s) ⇀ NH3(g) + H2S(g)
We can see that one mole of NH4HS produces one mole of NH3 and one mole of H2S. Therefore, we can write the equations:
[NH4HS] = [NH4HS]0 - x
and
[H2S] = y
Substituting these values back into the rearranged equilibrium constant equation:
[NH3] = (Kc * [NH4HS]) / [H2S]
[NH3] = (0.00016 * ([NH4HS]0 - x)) / y
Since we are assuming that NH4HS is completely consumed in the reaction and that NH3 and H2S are formed in a 1:1 ratio, we can substitute y with x:
[NH3] = (0.00016 * ([NH4HS]0 - x)) / x
Now, we can substitute the known values into the equation.
[NH3] = (0.00016 * (0 - x)) / x
[NH3] = -0.00016
Since we cannot have a negative concentration, the equilibrium concentration of NH3 is 0 mol/L.
Therefore, the equilibrium concentration of NH3 is 0 mol/L.