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4 years ago

Which analogy can best be likened to the activation energy of a chemical reaction?

1 answer:

6 0

The activation energy of a reaction is the minimum energy that must be overcome in order for the reaction to take place. One way of reaching the activation energy is by manipulating the process conditions like pressure or temperature. But the most common method is by adding an enzyme. An enzyme speeds up the rate of the reaction but does not actively take part in it.

An analogy would be pushing heavy wooden block down a slope. No matter how many people push on it, the block won't move because of friction. But if you spill oil on the floor, the block would effortlessly move down the slope. The oil here is like an enzyme in a reaction.

You might be interested in

What is the name of the salt with the structure K2S

gavmur [86]

Potassium sulphide :)

6 0

3 years ago

The half-life of radioactive element krypton-91 is 10 seconds. if 16 grams of krypton-91 are initially present, how many grams a

earnstyle [38]

10 seconds = 8grams
then just divide by 2 another 4 times...
= 0.5grams after 50 seconds

6 0

4 years ago

The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this

nevsk [136]

Answer:

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Explanation:

For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.

For a hypothetical reaction:

xA + yB ⇄ zC

The equilibrium constant is :

$Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }$

The given reaction involves the decomposition of H2O into H2 and O2

$2H_{2}O\rightleftharpoons 2H_{2} + O_{2}$

The equilibrium constant is expressed as :

$Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Since Keq = 5.31*10^-10

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

3 0

3 years ago

Read 2 more answers

How do trends in the periodic table help predict the properties of an element

irina1246 [14]

As you move around there is a change in: electronegativies, ionisation energies, atomic radius etc. different amounts of these properties are going to effect how the element acts

8 0

3 years ago

S8 + 24 F2 ⟶ 8 SF6

Arturiano [62]

Answer:

Theoretical Yield of SF₆ = 2.01 moles

Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.

Solution 1:

Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.

Given S₈ + 24F₂ => 8SF₆

425g 229g ?

= 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant

Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant.

S₈ = 1.66/1 = 1.66

F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant

Determining Theoretical Yield:

Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.

Given S₈ + 24F₂ => 8SF₆

425g 229g ?

= 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈ = 6.03 moles F₂ <= Limiting Reactant

Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈

Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆ to make the outcome smaller than 6.03. That is,

moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)

S₈ + 24F₂ => 8SF₆

moles SF₆ = 8/24(6.03) moles = 2.01 moles

You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.

This analysis works for all reaction stoichiometry problems.

Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.

____________________

Here's another example just for grins ...

C₂H₆O + 3O₂ => 2CO₂ + 3H₂O

Given: 253g 307g ? ?

a. Determine Limiting Reactant

b. Determine mass in grams of CO₂ & H₂O produced

Limiting Reactant

moles C₂H₆O = 253g/46g/mol = 5.5 moles => 5.5/1 = 5.5

moles O₂ = 307g/32g/mol = 9.6 moles => 9.6/24 = 0.4 ∴ O₂ is L.R.

But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.

C₂H₆O + 3O₂ => 2CO₂ + 3H₂O

------------ 9.6 mole (L.R.) ? ?

mole yield CO₂ = 2/3(9.6)mole = 6.4 mole (CO₂ coefficient < O₂ coefficient)

mole yield H₂O = 9.6mole = 9.6mole (coefficients O₂ & CO₂ are same.)

mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient C₂H₆O < coefficient O₂)

For grams => moles x formula weight (g/mole)

7 0

3 years ago