Question

Consider the reaction SO2Cl2(g) = SO2(g) + Cl2(g). The rate law for this reaction is rate...

Consider the reaction SO2Cl2(g) = SO2(g) + Cl2(g). The rate law for this reaction is rate = k[SO2Cl2] where k=2.90 ´ 10-4 s-1 at a particular temperature.

If the initial [SO2Cl2] = 0.0500 M, what is the initial rate?

What is the value of the half-life for this initial concentration of SO2Cl2 ?

What will be the half-life if the initial [SO2Cl2] = 0.0250 M

What will be the half-life if the initial [SO2Cl2] = 0.100 M

What will be the [SO2Cl2] after 2.00 minutes when we start with an initial [SO2Cl2] = 0.0500 M

How much time will it take for the [SO2Cl2] to decrease from 0.0500 M to 0.0250 M

How much time will it take for the [SO2Cl2] to decrease from 0.100 M to 0.0750 M

Homework Answers

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

t1/2 = 0.693 / K = 0.693 / 2.90 x 10-4

t1/2 = 2389.65 s

3) half is same for any initial concentration

t1/2 = 2389.65 s

4 ) t1/2 = 2389.65 s

5) for first order

K = (2.303 / t) log [A0] / [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

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The first order reaction, SO2Cl2 --> SO2 + Cl2, has a half life of 8.75 hours...
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 =...
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...
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...
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...
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...
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) +...
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...
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...
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...
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.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT