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valentinak56 [21]
3 years ago
8

The activation energy for the reaction no2(g)+co(g)⟶no(g)+co2(g) is ea = 100 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:
IrinaK [193]3 years ago
7 0
Answer: 350 kJ/mol

Explanation:

1) <span>δh = -250 kj/mol means that the products of the forward reaction are 250 kJ/mol lower in the potential enery diagram.

2) So, the reverse reaction must compensate the 250 kJ/mol and go upper as much as the activation energy of the forward reaction.

This is: Ea reverse = Ea forward + | </span><span>δh |

You can see this clearer on the attached graph of an exothermic reaction.

Ea reverse = 100 kJ / mol + | -250 kJ/mol | = 100 kJ/mol + 250 kJ/mol = 350 kJ/mol
</span>
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Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid])     (Eq. 01)

Where  

pKa = -log(Ka)        (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2    (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

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Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

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