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Vlada [557]
3 years ago
9

The formation of nitryl fluoride, 2no2(g)+f2(g)⇌2no2f(g) is first order in no2 and first order in f2. if the concentration of no

2 was increased by half and the concentration of f2 was quadrupled, by what factor would the reaction rate increase?
Chemistry
1 answer:
ra1l [238]3 years ago
4 0
The reaction is:

2 NO₂ (g) + F₂ (g) ⇆  2 NO₂F (g)

The stoichiometric coefficients of the substances balance out each other to obey the Law of Definite Proportions. Now, you have to note that determining the reaction rate expression is specific to a certain type of reaction. So, this are determined empirically through doing experiments. But in chemical reaction engineering, to make things simple, you assume that the reaction is elementary. This means that the order of a reaction with respect to a certain substance follows their individual stoichiometric coefficients. What I'm saying is, the stoichiometric coefficients are the basis of our reaction rate orders. For this reaction, the rate order is 2 for NO₂, 1 for F₂ and 2 for NO₂F. When the forward and reverse reactions are in equilibrium, then it applies that:

Reaction rate of disappearance of reactants = Reaction rate of formation of products.

Therefore, we can have two reaction rate constants for this. But since the conditions manipulated are the reactant side, let's find the expression for reaction rate of disappearance of reactants.

-r = k[NO₂]²[F₂]

The negative sign before r signifies the rate of disappearance. If it were in terms of the product, that would have been positive. The term k denotes for the reaction rate constant. That is also empirical. As you can notice the stoichiometric coefficients are exponents of the concentrations of the reactants. Let's say initially, there are 1 M of NO₂ and 1 M of F₂. Then,

-r = k(1)²(1)
-r = k

Now, if we change 1 M of NO₂ by increasing it to its half, it would now be 1.5 M NO₂. Then, if we quadruple the concentration of F₂, that would be 4 M F₂. Substituting the values:

-r = k(1.5)²(4)
-r = 9k

So, as you can see the reaction rate increase by a factor of 9.
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A sealed vessel containg helium, chlorine and fluorine has a total pressure of 205 kpa and partial pressures of 15 kpa for chlor
MrRa [10]

According to Dalton's law of partial pressure

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So if we know the partial pressure of all the gases we can determine the total pressure of the mxiture by adding all of them.

Thus

Ptotal = PCl2 + PF2 + PHe

Thus

Partial pressure of helium = 205 - ( 89 + 15) = 101 kPa

7 0
3 years ago
1. The
rodikova [14]

Answer:

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3 0
3 years ago
Given the reactions below, (1) Na(s) + H2O(l) → NaOH(s) + 1/2 H2(g) LaTeX: \Delta H^o_{rxn}Δ H r x n o = −146 kJ (2) Na2SO4(s) +
slamgirl [31]

Answer:

   ΔrxnH  = -580.5 kJ

Explanation:

To solve this question we are going to help ourselves with Hess´s law.

Basically the strategy here is to work  in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.

Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-

The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.

This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.

Taking  1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.

      Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l)             ΔrxnHº = 259 / 2 kJ  1/2 eq3

+    2NaOH(s)  + SO₃(g) ⇒  Na₂SO₄ (s) + H₂O (l)   ΔrxnHº = -418 kJ     - eq 2

+    2Na (s) + 2 H₂O (l)  ⇒   2 NaOH (s) + H₂ (g)    ΔrxnHº = -146 x 2    2 eq 1

<u>                                                                                                                                         </u>

Na₂O (s) + SO₃ (g)  ⇒ Na₂SO₄ (s)    ΔrxnHº =  259/2 + (-418) + (-146) x 2 kJ

                                                          ΔrxnH  = -580.5 kJ

4 0
3 years ago
You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2H
KIM [24]

Answer:

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+    pKa1 = 2.14

H2PO4 = HPO4 + H+       pKa2 = 6.86

HPO4 = PO4 + H+      pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

Download odt
7 0
3 years ago
How many grams are in 2.5 moles of N2? <br><br> (2 decimal places)
Leokris [45]

Answer: How many grams are in 2.5 moles of N2?

Explanation: 1 mole is equal to 1 moles N2, or 28.0134 grams.

 One mole of N2 molecules would have a mass of 2 X 14.01 g = 28.02 g.

3 0
3 years ago
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