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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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Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

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