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AleksAgata [21]
3 years ago
6

8. Select the lattice energy for rubidium chloride from the following data (in kJ/mol]

Chemistry
1 answer:
yKpoI14uk [10]3 years ago
4 0

Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

Lattice energy for rubidium chloride ( RbCl) is represented by the equation 6

Equation 1 represents the change in enthalpy for formation of RbCl

Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

So Equation 1 ≡  ΔH_{f} = -431 kJ/mol

Equation 2 ≡ Rb(s) ---> Rb(g) = 85.8  kJ/mol

Equation 3 ≡ IE1(Rb) = 397.5  kJ/mol

Equation 4 ≡ BE(Cl2) = 226  kJ/mol

Equation 5 ≡ Electron Affinity Cl = -332  kJ/mol

Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

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Morgarella [4.7K]

Answer:

The enthalpy of the reaction is 64.9 kJ/mol.

Explanation:

H_2 + \frac{1}{2}O_2\rightarrow H_2O,\Delta H_1 =-285.8 kJ..[1]

CH_4 + 2O_2\rightarrow CO_2 + 2H_2O,\Delta H_2 =-890.4 kJ..[2]

C_2H_6 + \frac{7}{2}O_2\rightarrow 2CO_2 + 3H_2O,\Delta H_3= -1559.9 kJ..[3]

2CH_4(g)\rightarrow C_2H_6(g) + H_2(g),\Delta H_4=?..[4]

2 × [2] - [1]- [3] = [4]  (Using Hess's law)

\Delta H_4=2\times \Delta H_2 -\Delta H_1 -\Delta H_3

\Delta H_4=2\times (-890.4 kJ)-(-285.8 kJ) -(-1559.9 kJ)

\Delta H_4=64.9 kJ/mol

The enthalpy of the reaction is 64.9 kJ/mol.

3 0
3 years ago
Hii pls help me to balance the equation thanksss​
tiny-mole [99]

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

\boxed{\pmb{\color{gold}{\sf{2SO_{2}(g) + O_{2}(g)\dashrightarrow 2SO_{3}(g)}}}}

▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓

7 0
3 years ago
Read 2 more answers
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1.00 x 10^-5 M

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