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tigry1 [53]
3 years ago
14

The decomposition of dinitrogen pentaoxide has an activation energy of 102 kJ/mol and ΔH°rxn = + 55 kJ/mol. What is the activati

on energy for the reverse reaction?
Chemistry
1 answer:
ahrayia [7]3 years ago
3 0

<u>Answer:</u> The activation energy of the reverse reaction is 47 kJ/mol

<u>Explanation:</u>

The chemical equation for the decomposition of dinitrogen pentaoxide follows:

N_2O_5\rightleftharpoons 2NO_2+\frac{1}{2}O_2

We are given:

Activation energy of the above reaction (forward reaction) = 102 kJ/mol

Enthalpy of the reaction = +55 kJ/mol

As, the enthalpy of the reaction is positive, the reaction is said to be endothermic in nature.

To calculate the activation energy for the reverse reaction, we use the equation:

E_a_{\text{(forward)}}=E_a_{\text{(backward)}}+|\Delta H|

where,

E_a_{\text{(forward)}} = Activation energy of the forward reaction = 102 kJ/mol

E_a_{\text{(backward)}} = Activation energy of the backward reaction = ?

\Delta H = Enthalpy of the reaction = +55 kJ/mol

Putting values in above equation, we get:

102=E_a_{\text{(backward)}}+55\\\\E_a_{\text{(backward)}}=(102-55)=+47kJ/mol

Hence, the activation energy of the reverse reaction is 47 kJ/mol

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a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh
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Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

Molarity = M = \frac{1,22 mol solute}{lts solution}

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution

With the molecular weight of solute (<em>Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol</em>), we can obtain the mass of solute:

1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: <em>Moles solution = moles solute + moles solvent.</em> First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent  

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03

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