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Amiraneli [1.4K]
3 years ago
14

A certain first-order reaction has a half-life of 20.0 minutes. How much time is required for this reaction to be 75% complete?

Chemistry
1 answer:
ryzh [129]3 years ago
3 0

Answer:

X = 2

Explanation:

As you know, the rate of a first-order reaction depends linearly on the concentration of a single reactant. The rate of a first-order reaction that takes the form

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Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is c
oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be P_1

and the total pressure at the final equilibrium be P_2

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

P_1V_1 = P_2V_2

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

V_2 =  \dfrac{V_1}{\dfrac{3}{2}} ----- (1)

also;

Thus ;

P_1V_1 = P_2(  \dfrac{V_1}{\frac{3}{2}})

P_1V_1 = P_2 * 2  \dfrac{V_1}{3}

3 P_1 V_1 = 2 P_2 V_1

Dividing both sides by V_1

3P_1 = 2P_2

P_2 =P_1 \dfrac{3}{2}  ----- (2)

From ;

P_1V_1 = P_2V_2

P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}

P_2 V_2 = P_1 * \dfrac{3}{2}*   \dfrac{2 }{3}}*V_1

P_2 V_2 = P_1 V_1

Thus; The new equilibrium total pressure will be  increased to one-half to initial total pressure.

7 0
3 years ago
How far is Louisville from the nearest oil field?​
rjkz [21]

Answer:

anyywhere between 20-23 miles

Explanation:

it depends what type of oil field u mean

7 0
3 years ago
Read 2 more answers
Energy cannot be created or destroyed. Does the statement describe a scientific law?
LekaFEV [45]
Yes, the law of thermodynamics, or Law of Conservation of Energy
3 0
3 years ago
What is the stoichiometric coefficient for oxygen when the following equation is balanced using the lowest whole-number coeffici
rodikova [14]

Answer:

Option (B) 7

Explanation:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

To know the coefficient of O2 in the above equation, let us balance the equation.

The above equation can be balance as follow:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

There are 3 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 3 in front of CO2 as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + H2O(l)

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + 3H2O(l)

There are a total of 4 atoms of O on the left side and a total of 9 atoms on the right side. It can be balance by putting 7/2 in front of O2 as show below:

C3H6O2(l) + 7/2O2(g) → 3CO2(g) + 3H2O(l)

Multiply through by 2

2C3H6O2(l) + 7O2(g) → 6CO2(g) + 6H2O(l)

Now, the equation is balanced.

From the balanced equation above, the coefficient of O2 is 7.

7 0
3 years ago
The energy transformations are similar because they both involve transformations that
blondinia [14]
Sorry but I don’t understand the question. Sorry I’m not any help
3 0
2 years ago
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