Question

Consider the reaction SO_{2}Cl_{2}(g) =
SO_{2}(g) + Cl_{2}(g). The rate law for this
reaction is rate = k[SO_{2}Cl_{2}] where k=2.90 ´
10^{-4} s^{-1} at a particular temperature.

If the initial [SO_{2}Cl_{2}] = 0.0500 M, what
is the initial rate?

What is the value of the half-life for this initial
concentration of SO_{2}Cl_{2} ?

What will be the half-life if the initial
[SO_{2}Cl_{2}] = 0.0250 M

What will be the half-life if the initial
[SO_{2}Cl_{2}] = 0.100 M

What will be the [SO_{2}Cl_{2}] after 2.00
minutes when we start with an initial
[SO_{2}Cl_{2}] = 0.0500 M

How much time will it take for the
[SO_{2}Cl_{2}] to decrease from 0.0500 M to 0.0250
M

How much time will it take for the
[SO_{2}Cl_{2}] to decrease from 0.100 M to 0.0750
M

Answer #1

rate = k[SO2Cl2]

1) if [SO2Cl2] = 0.0500 M

rate = 2.90 x 10^{-4} [0.0500]

rate = 1.45 x 10^{-5} M/s

2) for first order reaction

t_{1/2} = 0.693 / K = 0.693 / 2.90 x 10^{-4}

**t _{1/2} = 2389.65 s**

3) half is same for any initial concentration

t_{1/2} = 2389.65 s

4 ) t_{1/2} = 2389.65 s

5) for first order

K = (2.303 / t) log [A_{0}] / [A]

2.90 x 10^{-4} = (2.303 / 2.0 x60) log [0.05] / [A]

log [0.05] / [A] = 0.0151

-1.30 - log [A ] = 0.0151

log [A] = -1.3151

**[A] = 0.0484 M**

6) 2.90 x 10^{-4} = (2.303 / t) log [0.05] /[0.025]

2.90 x 10^{-4} = (2.303 /t) 0.301

2.303/ t = 0.0009634

**t = 2390.5 s**

**7)** 2.90 x 10^{-4} = (2.303 /t) log
[0.1] /[0.075]

2.90 x 10^{-4} = (2.303 /t) 0.125

2.303 /t = 0.00232

**t = 992.67 s**

The first order reaction, SO2Cl2 --> SO2 + Cl2, has a half
life of 8.75 hours at 593 K.?
How long will it take for the concentration of SO2CL2 to fall to
12.5% of its initial value?

13. Consider the following equilibrium reaction:
SO2Cl2 (g) ⇌ Cl2 (g) + SO2 (g) Kc = 2.99 x 10-7
@227o C
If the initial concentration of SO2Cl2 is 0.1680 M, determine
the equilibrium concentrations of SO2Cl2, Cl2, and SO2.
I know that this type of problem requires an ICE chart but im
not too sure how to solve it.

Consider the following reaction:
SO2Cl2(g)⇌SO2(g)+Cl2(g)
A reaction mixture is made containing an initial [SO2Cl2] of
2.2×10−2 M . At equilibrium, [Cl2]=1.0×10−2 M
.
Calculate the value of the equilibrium constant
(Kc).
Express your answer using two significant figures.

Consider the reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) Kp=2.91×103 at
298 K In a reaction at equilibrium, the partial pressure of SO2 is
159 torr and that of Cl2 is 285 torr . Part A What is the partial
pressure of SO2Cl2 in this mixture?

Consider the reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) Kp=2.91×103 at
298 K In a reaction at equilibrium, the partial pressure of SO2 is
144 torr and that of Cl2 is 287 torr .
Part A What is the partial pressure of SO2Cl2 in this
mixture?

For the reaction SO2 (g) + Cl2 (g) ⇔ SO2Cl2 (g), Kc = 55.5 at a
certain temperature.
If 1.00 mole of SO2 (g) and 1.00 mole of Cl2 (g) are placed in a
10.0-L container and allowed to
come to equilibrium, what is the equilibrium concentration of
SO2Cl2?

Sulfuryl chloride decomposes to sulfur dioxide and chlorine by
the following reaction: SO2Cl2(g) ⇌ SO2(g) + Cl2(g) Kc = 0.045 at
648 K If an initial concentration of 6.76 x 10-2 M SO2Cl2 is
allowed to equilibrate, what is the equilibrium concentration of
Cl2?

Consider the following reaction:
SO2Cl2(g)⇌SO2(g)+Cl2(g)
Kc=2.99×10−7 at 227∘C
Part A
If a reaction mixture initially contains 0.159 M SO2Cl2, what is
the equilibrium concentration of Cl2 at 227 ∘C?

Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g)
Kc=2.99×10−7 at 227∘C
If a reaction mixture initially contains 0.195 M SO2Cl2, what is
the equilibrium concentration of Cl2 at 227 ∘C?

At 100 ∘C, Kc=0.078 for the reaction SO2Cl2(g)←−→SO2(g)+Cl2(g)
In an equilibrium mixture of three gases, the concentrations of
SO2Cl2 2 and SO2 are 0.130 M and 0.125 M respectively.
Part A What is the partial pressure of Cl2 in the equilibrium
mixture?
Express your answer using two significant figures.

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