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drek231 [11]
3 years ago
15

Consider this reaction:2HI(g) ------> H2(g) + I2(g)At a certain temperature it obeys this rate law.rate = (8.74 x 10^-4s^-1)

[HI] 1. Suppose a vessel contains HI at a concentration of 0.330M . Calculate the concentration of HI in the vessel 800 seconds later. You may assume no other reaction is important.
Chemistry
1 answer:
nasty-shy [4]3 years ago
8 0

Answer:

0.1440M

Explanation:

Let''s bring out the parameters we were given;

Rate constant = 8.74 x 10^-4s^-1

Initial Concentration [A]o = 0.330M

Final concentration [A]= ?

Time = 800s

The reaction is a first order reaction, due to the unit of the rate constant. In first order reactions, the reaction rate is directly proportional to the reactant concentration and the units of first order rate constants are 1/sec.

Formular relating these parameters is given as;

ln[A] = ln[A]o − kt

Making [A] subject of interest, we have;

ln[A] = ln[A]o − kt

ln[A]  = ln(0.330) - ( 8.74 x 10^-4 * 800)

In[A] = - 1.1086 - (6992 x 10^-4)

ln[A] = -1.8079

[A] = 0.1440M

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Answer:

\boxed{\text{ B. Increase the temperature and decrease the pressure.}}

Explanation:

Let's say the reaction is

R ⇌ 2P; endothermic

I like to consider heat as if it were a reactant or a product in a chemical equilibrium.

Another way to write the equilibrium would be

heat + R ⇌ 2P

According to Le Châtelier's Principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Let's consider each of the stresses in turn.

(i) Changing the temperature

If you want to increase the amount of product, you increase the temperature. The system will try to get rid of the added heat by shifting to the right, thus forming more product.

(ii) Changing the pressure

If R and P are liquids or solids or in aqueous solution, changing the pressure will have no effect. Something must be in the gas phase for a change in pressure to affect the position of equilibrium.

If P is a gas, the equilibrium is

heat + R ⇌ 2P(g)

Then, decreasing the pressure will produce more P. If you reduce the pressure, the system will respond by shifting to the right (the side with more gas molecules) to produce more P and bring the pressure back up

5 0
3 years ago
The force of a gas's outward push divided by the area of the walls of the container is the gas's
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The temperature of a 100.0 g sample of water is raised from 30degrees celsius to 100.0 degrees celsius. How much energy is requi
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Answer:

29260J

Explanation:

Given parameters:

Mass of water sample  = 100g

Initial temperature = 30°C

Final temperature  = 100°C

Unknown:

Energy required for the temperature change = ?

Solution:

The amount of heat required for this temperature change can be derived from the expression below;

     H  = m c (ΔT)

H is the amount of heat energy

m is the mass

c is the specific heat capacity of water  = 4.18J/g°C

ΔT is the change in temperature

Now insert the parameters and solve;

          H  = 100 x 4.18 x (100 - 30)

          H  = 100 x 4.18 x 70 = 29260J

6 0
2 years ago
What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0
3 years ago
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