Write the equilibrium expression for the following reactions:
a. wW +xX ⇌ uU +vV
b. H2O(g)+CO(g)⇌ H2 (g)+CO2(g )
c. COCl2(g )⇌CO(g )+Cl2( g)
d. H2(g)+Cl2(g)⇌ 2HCl(g)+energy
e. CO(g )+ NO2(g)⇌ NO(g )+CO2 (g)
f. Zn(s)+2 Ag(aq) ^? (aq) = Zn ^2+? (aq) + 2Ag (s) ??
g. C2H6(g)⇌ H2 (g)+C2H 4(g)
2. Consider the following equilibrium reaction:
H2(g)+Br2 (g)⇌2 HBr(g )+energy
The Keq for this reaction at 25 is 1.02. At equilibrium the concentration of HBr is ℃
0.50mol/L. Assuming H2 and Br2 are present in equal amounts, calculate the
concentration of H2 at equilibrium.
3. Analysis of the following equilibrium reaction at 900 provides the concentrations ℃
listed below:
H2O(g)+CO(g)⇌ H2 (g)+CO2(g )
Experiment [H2O] [CO] [H2] [CO2]
1 0.352 0.352 0.148 0.648
2 0.266 0.266 0.234 0.234
3 0.686 0.186 0.314 0.314
Write the equilibrium expression for the reaction and calculate the value of the
equilibrium constant for each experiment.
4. In the following reaction at 448 , the equilibrium concentrations are HI = 0.0040M, ℃
H2 = 0.0075M, I2 = 0.000043M. Calculate the equilibrium constant given the reaction
below:
2 HI(g)⇌ H2 (g)+I 2(g )
5. If the temperature of an exothermic reaction at equilibrium is lowered, is the value of
Keq increased or decreased
a. K = [U]^u[V]^v / [W]^w[X]^x
b. K = [H2]^1[CO2]^1 / [H2O]^1[CO]^1
c. K = [CO]^1[Cl2]^1 / [COCl2]^1
d. K = [HCl]^2 / [H2]^1[Cl2]^1
e. K = [NO]^1[CO2]^1 / [CO]^1[NO2]^1
f. K = [Zn^2+]?^a[Ag]^2 / [Zn]?^b[Ag]^2
g. K = [H2]^1[C2H4]^1 / [C2H6]^1[H2]^1
2. The concentration of H2 at equilibrium can be calculated using the expression:
Keq = ([HBr]^2) / ([H2][Br2])
Since [HBr] = 0.50 M, we can substitute this value into the expression:
1.02 = (0.50^2) / ([H2][0.50])
Simplifying, we get:
[H2] = (0.50^2) / (1.02*0.50) = 0.490 M
Therefore, the concentration of H2 at equilibrium is 0.490 M.
3. The equilibrium expression for the reaction is:
K = [H2][CO2] / [H2O][CO]
For each experiment, we can substitute the given concentrations into the expression to calculate the equilibrium constant:
Experiment 1: K = (0.148)(0.648) / (0.352)(0.352) = 0.244
Experiment 2: K = (0.234)(0.234) / (0.266)(0.266) = 0.231
Experiment 3: K = (0.314)(0.314) / (0.686)(0.186) = 0.256
4. The equilibrium constant can be calculated using the expression:
K = ([H2][I2]) / ([HI]^2)
Substituting the given concentrations:
K = (0.0075)(0.000043) / (0.0040^2) = 0.427 M^-1
Therefore, the equilibrium constant is 0.427 M^-1.
5. If the temperature of an exothermic reaction at equilibrium is lowered, the value of Keq is increased. This is because the forward reaction is exothermic, meaning it releases heat. By lowering the temperature, the system tries to counteract the decrease in temperature by shifting towards the side of the reaction that produces heat, which is the forward reaction. This leads to an increase in the concentration of the reactants and a decrease in the concentration of the products, resulting in an increased value of Keq.
The answer I got for question 2 is [H2] = 0.50 mol/L.
The answer I got for question 3 is
K =([H2] [CO2]) / ([H2O] [CO])
Expt 1: K = 0.774
Expt 2: K = 0.774
Expt 3: K = 0.773
The answer I got for question 4 is K= 0.020
Can you show me the steps?
Apologies for the incorrect response. Let's revisit the calculations for questions 2, 3, and 4.
2. To find the concentration of H2 at equilibrium, we can use the equilibrium expression:
Keq = ([HBr]^2) / ([H2][Br2])
Given that [HBr] = 0.50 M, and assuming H2 and Br2 are present in equal amounts, let's assign the concentration of H2 at equilibrium as x. Therefore, [Br2] would also be x.
Substituting these values into the equilibrium expression, we have:
1.02 = (0.50^2) / (x*x)
Multiply both sides by x^2:
1.02x^2 = 0.50^2
Solving for x:
x^2 = (0.50^2) / 1.02
x^2 ≈ 0.245
Taking the square root of both sides:
x ≈ √(0.245)
x ≈ 0.495 M
Therefore, the concentration of H2 at equilibrium is approximately 0.495 M.
3. Let's calculate the equilibrium constants for each experiment using the given concentrations and the following equilibrium expression:
K = ([H2] [CO2]) / ([H2O] [CO])
Experiment 1:
K = (0.148)(0.648) / (0.352)(0.352) ≈ 0.790
Experiment 2:
K = (0.234)(0.234) / (0.266)(0.266) ≈ 0.780
Experiment 3:
K = (0.314)(0.314) / (0.686)(0.186) ≈ 0.799
Therefore, the equilibrium constants for experiments 1, 2, and 3 are approximately 0.790, 0.780, and 0.799, respectively.
4. To calculate the equilibrium constant for the given reaction, we can use the equilibrium expression:
K = ([H2][I2]) / ([HI]^2)
Given the equilibrium concentrations as follows:
[HI] = 0.0040 M
[H2] = 0.0075 M
[I2] = 0.000043 M
Plugging these values into the equilibrium expression, we have:
K = (0.0075)(0.000043) / (0.0040^2)
K = 0.000025575 / 0.000016 = 1.60
Therefore, the equilibrium constant for the reaction is 1.60.
For question 2, I got [H2] = 0.50 mol/L. how did you get 0.495 M?
Apologies for the confusion. Let's go through the calculation for question 2 again.
Given that at equilibrium [HBr] = 0.50 M, and assuming [H2] and [Br2] are present in equal amounts, we can assign the concentration of [H2] as x. Therefore, [Br2] would also be x.
The equilibrium expression is: Keq = ([HBr]^2) / ([H2][Br2])
Substitute the given values: 1.02 = (0.50^2) / (x * x)
Rearrange the equation: (x * x) = (0.50^2) / 1.02
Simplify: x^2 = (0.50^2) / 1.02
Solve for x by taking the square root of both sides: x = sqrt((0.50^2) / 1.02)
Calculating the value: x ≈ 0.50 mol/L
Therefore, the concentration of H2 at equilibrium is approximately 0.50 mol/L.