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bogdanovich [222]
1 year ago
10

The activation energy for the reaction NO2(g)+CO(g)⟶NO(g)+CO2(g) is Ea = 375 kJ/mol and the change in enthalpy for the reaction

is ΔH = -250 kJ/mol . What is the activation energy for the reverse reaction?
Chemistry
1 answer:
HACTEHA [7]1 year ago
4 0
Answer: 625 kj/mol

Explanation:

As shown below this expression gives the activation energy of the reverse reaction:

EA reverse reaction = EA forward reaction + | enthalpy change |

1) The activation energy, EA is the difference between the potential energies of the reactants and the transition state:

EA = energy of the transition state - energy of the reactants.

2) The activation energy of the forward reaction given is:

EA = energy of the transition state - energy of  [ NO2(g) + CO(g) ] = 75 kj/mol

3) The negative enthalpy change - 250 kj / mol for the forward reaction means that the products are below in the potential energy diagram, and that the potential energy of the products, [NO(g) + CO2(g) ] is equal to 375 kj / mol - 250 kj / mol = 125 kj/mol

4) For the reverse reaction the reactants are [NO(g) + CO2(g)], and the transition state is the same than that for the forward reaction.

5) The difference of energy between the transition state and the potential energy of [NO(g) + CO2(g) ] will be the absolute value of the change of enthalpy plus the activation energy for the forward reaction:

EA reverse reaction = EA forward reaction + | enthalpy change |

EA reverse reaction = 375 kj / mol + |-250 kj/mol | = 375 kj/mol + 250 kj/mol = 625 kj/mol.

And that is the answer, 625 kj/mol
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Answer:

The answer to your question is: 6 moles of HNO₃

Explanation:

Data

Volume = 25 ml

Concentration = 6 M HNO₃

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Formula

Molarity = # moles / volume

# of moles = Volume x Molarity

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# of moles = 0.10 x 6

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6 0
2 years ago
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A 0.580 g sample of a compound containing only carbon and hydrogen contains 0.480 g of carbon and 0.100 g of hydrogen. At STP, 3
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Answer:

Molecular formula for the gas is: C₄H₁₀

Explanation:

Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g

At STP → 1 atm and 273.15K

1 atm . 0.0336 L = n . 0.082 . 273.15 K

n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)

n = 1.500 × 10⁻³ moles

Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m

Now we propose rules of three:

If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C

58 g of gas (1mol) would have:

(58 g . 0.480) / 0.580 = 48 g of C  

(58 g . 0.100) / 0.580 = 10 g of H

 48 g of C / 12 g/mol = 4 mol

 10 g of H / 1g/mol = 10 moles

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If the manta ray gets 50 kcal of energy by eating the starfish, how much energy does the clam get from eating the plankton?
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Compound X has a molar mass of 104.01·gmol−1 and the following composition: element mass % nitrogen 26.93% fluorine 73.07% Write
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Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f
Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0
2 years ago
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