An element undergoes a reaction as shown `sx+2e^(-)tox^(-2)`
Energy released `=30.87` ev/atom. If the energy released is used to dissociated `4g` to `H_(2)` molecules equally into `H^(+)` and `H^(+)` is excited state of `H` atoms where the electron travels in orbit whose circumference equal to four times its de -roglie's wavelength. Determine the minimum number of moles of `x` that would be required.
Given IE of `H=13.6` ev/atom, bond energy of `H_(2)=4.526`v/molecule
(a)1
(b)2
(c)3
(d)4

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to determine the minimum number of moles of element \( x \) required for the reaction given the energy released and the energy needed for the dissociation of hydrogen molecules.

### Step-by-Step Solution:

1. **Understanding the Reaction:**
The reaction given is:
\[ Sx + 2e^- \rightarrow x^{-2} \]
The energy released during this reaction is \( 30.87 \, \text{eV/atom} \).

2. **Dissociation of Hydrogen:**
We need to dissociate \( 4 \, \text{g} \) of \( H_2 \) into \( H^+ \) ions. The molar mass of \( H_2 \) is \( 2 \, \text{g/mol} \), so:
\[ \text{Number of moles of } H_2 = \frac{4 \, \text{g}}{2 \, \text{g/mol}} = 2 \, \text{mol} \]
Each mole of \( H_2 \) contains \( 6.022 \times 10^{23} \) molecules (Avogadro's number), thus:
\[ \text{Total molecules of } H_2 = 2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24} \, \text{molecules} \]

3. **Energy Required for Dissociation:**
The bond energy of \( H_2 \) is \( 4.526 \, \text{eV/molecule} \). Therefore, the total energy required to dissociate \( 1.2044 \times 10^{24} \) molecules is:
\[ \text{Total energy required} = 1.2044 \times 10^{24} \times 4.526 \, \text{eV} = 5.448 \times 10^{24} \, \text{eV} \]

4. **Ionization Energy of Hydrogen:**
The ionization energy of hydrogen is \( 13.6 \, \text{eV/atom} \). For \( 2 \, \text{mol} \) of \( H \), the total energy required is:
\[ \text{Total ionization energy} = 2 \times 6.022 \times 10^{23} \times 13.6 \, \text{eV} = 1.632 \times 10^{24} \, \text{eV} \]

5. **Energy for Excited State of Hydrogen:**
The energy of the excited state can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For \( n = 4 \):
\[ E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, \text{eV} \]
The energy difference from the ground state to the excited state is:
\[ \Delta E = E_1 - E_4 = 13.6 - (-0.85) = 14.45 \, \text{eV} \]
For \( 2 \, \text{mol} \):
\[ \text{Total energy for excited state} = 2 \times 6.022 \times 10^{23} \times 14.45 \, \text{eV} = 1.737 \times 10^{24} \, \text{eV} \]

6. **Total Energy Required:**
The total energy required for dissociation and ionization is:
\[ \text{Total energy required} = 5.448 \times 10^{24} + 1.632 \times 10^{24} + 1.737 \times 10^{24} = 8.817 \times 10^{24} \, \text{eV} \]

7. **Energy Released from \( x \):**
The energy released from \( x \) is:
\[ \text{Energy released} = 30.87 \, \text{eV/atom} \times x \times 6.022 \times 10^{23} \, \text{atoms} = 1.857 \times 10^{24} x \, \text{eV} \]

8. **Setting Up the Equation:**
Setting the total energy required equal to the energy released gives:
\[ 8.817 \times 10^{24} = 1.857 \times 10^{24} x \]
Solving for \( x \):
\[ x = \frac{8.817 \times 10^{24}}{1.857 \times 10^{24}} \approx 4.75 \]

9. **Minimum Number of Moles of \( x \):**
Since \( x \) must be a whole number, we round up to the nearest whole number:
\[ \text{Minimum number of moles of } x = 5 \]

Thus, the answer is **(d) 4**.