What kind of chemical reaction is this?

Identify the catalyst in this reaction, explain how

telo118 [61]

Question:

Sulfuric acid was once produced through the reaction of sulfur trioxide with water. Sulfur trioxide can form through the reaction of sulfur dioxide and oxygen gas. When nitrogen monoxide gas is added to the system, the reaction speeds up significantly because it proceeds through the following steps:

equations

Identify the catalyst in this reaction, explain how you know it is the catalyst, and describe how it increases the rate of the reaction.

Answer:

NO

It is present but not consumed

NO Lowers the activation energy of the reaction

Explanation:

A catalyst is a substance that is present in a chemical reaction and enables the reaction to occur at a faster rte but does not take part n the reaction

Therefore, whereby NO is not consumed, it is the catalyst

It functions by lowering the activation energy

5 0

3 years ago

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch

mamaluj [8]

Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

$2SO_2 + O_2 \to 2SO_3$

The I.C.E table can be represented as:

2SO₂ O₂ 2SO₃

Initial: 14 2.6 0

Change: -2x -x +2x

Equilibrium: 14 - 2x 2.6 - x 2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction $2SO_2 + O_2 \to 2SO_3$ = 0.8325;

If we want to find:

$SO_2 + \dfrac{1}{2}O_2 \to SO_3$

Then:

$K_c = (0.8325)^{1/2}$

$\mathbf{K_c = 0.912}$

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

7 0

3 years ago

What effects does the concentration of reactants have on the rate of a reaction ?

Harlamova29_29 [7]

Question:

What effects does the concentration of reactants have on the rate of a reaction?

Answer:

Reactant concentration. Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.

Increasing the concentration of reactants generally increases the rate of reaction because more of the reacting molecules or ions are present to form the reaction products. ... When concentrations are already high, a limit is often reached where increasing the concentration has little effect on the rate of reaction.

Hope this helps, have a good day. c;

5 0

3 years ago

Given the formula representing a compound: what is a chemical name of this compound? 2-pentene 2-pentyne 3-pentene 3-pentyne

Aleks [24]

The question is incomplete. Complete question is attached below.
..........................................................................................................................

Correct Answer: Option 1) 2-pentene

Reason:
Following are the IUPAC rules for naming the compound
1) Select the longest carbon chain. In present case longest carbon chain has 5 carbon atom. Hence, it is a pentane derivative.
2) In case of alkene, replace 'e' of alkane by 'ene'
3) Give lowest number to function group. In present case, it is double bond.

Applying above rules, the IUPAC name of compound is 2-pentene

4 0

3 years ago

Consider the titration of 1L of 0.36 M NH3 (Kb=1.8x10−5) with 0.74 M HCl. What is the pH at the equivalence point of the titrati

worty [1.4K]

Answer:

C

Explanation:

The question asks to calculate the pH at equivalence point of the titration between ammonia and hydrochloric acid

Firstly, we write the equation of reaction between ammonia and hydrochloric acid.

NH3(aq)+HCl(aq)→NH4Cl(aq)

Ionically:

HCl + NH3 ---> NH4 + Cl-

Firstly, we calculate the number of moles of the ammonia as follows:

from c = n/v and thus, n = cv = 0.36 × 1 = 0.36 moles

At the equivalence point, there is equal number of moles of ammonia and HCl.

Hence, volume of HCl = number of moles/molarity of HCl = 0.36/0.74 = 0.486L

Hence, the total volume of solution will be 1 + 0.486 = 1.486L

Now, we calculate the concentration of the ammonium ions = 0.36/1.486 = 0.242M

An ICE TABLE IS USED TO FIND THE CONCENTRATION OF THE HYDROXONIUM ION(H3O+). ICE STANDS FOR INITIAL, CHANGE AND EQUILIBRIUM.

NH4+ H2O ⇄ NH3 H3O+

I 0.242 0 0

C -X +x +X

E 0.242-X X X

Since the question provides us with the base dissociation constant value K b, we can calculate the acid dissociation constant value Ka

To find this, we use the mathematical equation below

K a ⋅ K b = K w

, where K w- the self-ionization constant of water, equal to

10 ^-14 at room temperature

This means that you have

K a = K w.K b = 10 ^− 14 /1.8 * 10^-5 = 5.56 * 10^-10

Ka = [NH3][H3O+]/[NH4+]

= x * x/(0.242-x)

Since the value of Ka is small, we can say that 0.242-x ≈ 0.242

Hence, K a = x^2/0.242 = 5.56 * 10^-10

x^2 = 0.242 * 5.56 * 10^-10 = 1.35 * 10^-10

x = 0.00001161895

[H3O+] = 0.00001161895

pH = -log[H3O+]

pH = -log[0.00001161895 ] = 4.94

7 0

3 years ago