1.The following data was obtained for the reaction at 25°C: BrO3– (aq) + 5Br– (a
ID: 876597 • Letter: 1
Question
1.The following data was obtained for the reaction at 25°C:
BrO3–(aq) + 5Br–(aq) + 6H+(aq) 3Br2(aq) +3H2O(l) Exp
[BrO3–]0 (M)
[Br–]0 (M)
[H+]0 (M)
Initial Rate (M/min)
1
0.10
0.10
0.10
8.0 × 10–4
2
0.20
0.10
0.10
1.6 × 10–3
3
0.40
0.15
0.10
4.8 × 10–3
4
0.15
0.20
0.20
9.6 × 10–3
A. Determine the rate law for the reaction?
B. Determine the rate constant for the reaction?
c. What is the initial rate when [BrO3–]0 = 0.10 M, [Br–]0 = 0.50 M, and [H+]0 = 0.60 M?
D.. Adding a catalyst decreases the activation energy of the reaction by 25 kJ/mol. What is the rate constant for the catalyzed reaction?
2.The rate of the reaction shown below is known to be first order in I2 and first order in H2.
H2(g) + I2(g) 2HI(g)
The following mechanism was proposed for this reaction. Is it a valid mechanism? For full credit, you must determine the rate law as predicted by the mechanism, identify all intermediates, and provide a complete evaluation of all criteria necessary for the mechanism to be valid.
Step 1: I2(g) 2I(g) (fast, equilibrium)
Step 2: H2(g) + I(g) H2I(g) (fast, equilibrium)
Step 3: H2I(g) + I(g) 2HI(g) (slow)
3. When 9.25 g of ClF3 was introduced into an empty 2.00-L container at 700.0 K, 40.0% of the ClF3 decomposed to give an equilibrium mixture of ClF3, ClF, and F2. An additional 5.45 g of ClF was then added to the container. What are the molar concentrations of ClF3, ClF, and F2 when the mixture reaches equilibrium?
ClF3(g) ClF(g) + F2(g)
4. A 1.00-L vessel at 298 K is filled with 11.0 g of NOBr. How many grams of NO are in the vessel at equilibrium?
2NOBr(g) 2NO(g) + Br2(g) Kp = 4.00 × 10–5 at 298 K
Answer:______________________________
BrO3–(aq) + 5Br–(aq) + 6H+(aq) 3Br2(aq) +3H2O(l) Exp
[BrO3–]0 (M)
[Br–]0 (M)
[H+]0 (M)
Initial Rate (M/min)
1
0.10
0.10
0.10
8.0 × 10–4
2
0.20
0.10
0.10
1.6 × 10–3
3
0.40
0.15
0.10
4.8 × 10–3
4
0.15
0.20
0.20
9.6 × 10–3
Explanation / Answer
1) rate = k [BrO3-]x [Br-]y [H+]z
we need to find out x , y , z
From the table
by using data 1 & 2 we get x = 1
by using data 2 & 3 we get y = 1
by using data 3 & 4 we get z = 2
A) Rate law :
Rate = k [BrO3-]x [Br-]y [H+]z
Rate = k [BrO3-]1 [Br-]1[H+]2
B) Rate constant for the reaction :
Rate = k [BrO3-]1 [Br-]1[H+]2
8.0 104 = k [0.1]1 [0.1]1[0.1]2
k = 8 mol-3 lit3 sec-1
C) initial rate
[BrO3] = 0.10 M, [Br] = 0.50 M, and [H+] = 0.60 M
Rate = 8.0 104 M/min
Rate = k [BrO3-]1 [Br-]1[H+]2
= 8 [0.1]1 [0.5]1[0.6]2
Rate = 0.144
D) activation energy decreases rate of reaction increases
2)
yes . it is valid mechanism
rate = k [H2][I2]
explanation :
forward rate = k2[I][H2]
reverse rate = k-2[IH2]
k2[I][H2] = k-2[IH2]
so [IH2] = (k2 / k-2)[I][H2]
From the first fast equilibrium step, k1[I2] = k-1[I]^2
[I] = sqrt ((k1 / k-1) [I2])
Substitute the new expressions for [IH2] and [I] into our first rate law.
rate = k3[IH2][I] = k ((k2 / k-2)[I][H2]) (sqrt ((k1 / k-1) [I2]))
lump all of the k's into one constant, K4
rate = k4[I][H2](sqrt [I2]) = k4(sqrt ((k1 / k-1) [I2])[H2](sqrt [I2])
rate= k5[H2][I2]
3)
ClF3 molarity = (9.25 / 92.44 ) x 1/2
= 0.05M
ClF3 <------------------> ClF + F2
0.05 0 0
0.05-x x x
ClF molarity =(5.45 / 54.5 ) x1/2
=0.05 M
new equilibrium
ClF3 <------------------> ClF + F2
0.05 0 0
0.05-x x +0.05 x -------------------> new equilibrium
x = 0.05 x 40 / 100
= 0.02
at equilibrium concentrations
[ClF3] = 0.05-0.02 = 0.03 M
[ClF] = x + 0.05 = 0.07 M
[F2] = x=0.02 M
4)
NOBr molarity = 11/ 110 = 0.1
Kp = Kc (RT)^Dn
4 x 10^-5 = Kc (0.0821 x 298)^1
Kc = 1.635 x 10^-6
2 NOBr < -----------------------> 2NO + Br2
0.1 0 0 ----------------------initial
0.1-2x 2x x -------------------equilibrium
Kc = (2x)^2 (x) / (0.1-2x)^2
1.635 x 10^-6 = 4x^3/ (0.1-2x)^2
x = 1.565 x 10^-3
[NO] = 2x = 2 x 1.565 x 10^-3
molarity of [NO] = 3.13 x 10^-3 M
Molarity = n /V
3.13 x 10^-3 = n/1
n = 3.13 x 10^-3
(w/mw) = 3.13 x 10^-3
w/ 30 = 3.13 x 10^-3
w = 0.094 g
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