Question

(II) Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.110 nm, and their final separation is 0.100 nm. How much electric potential energy was lost in this reaction (in units of eV)?

Short answer

The electric potential energy lost during the reaction is \(1.25\;{\rm{eV}}\).

Step-by-step solution

Understanding of electric potential energy

The electric potential energy may be defined as the energy associated with a charge due to its location in an electric field. It is considered a component of mechanical energy.

Given information

The initial separation is,\({r_1} = 0.110\;{\rm{nm}}\).

The final separation is,\({r_2} = 0.100\;{\rm{nm}}\).

The charge on the electron is,\({q_1} = - 1.6 \times {10^{ - 19}}\;{\rm{C}}\)

The charge on the proton is,\({q_2} = 1.6 \times {10^{ - 19}}\;{\rm{C}}\)

Evaluation of the initial electric potential energy

The initial potential energy can be calculated as:

\(\begin{aligned}P{E_1} &= \frac{{k{q_1}{q_2}}}{{{r_1}}}\\ &= \frac{{\left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right)\left( { - 1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}}{{\left( {0.110\;{\rm{nm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{nm}}}}} \right)}}\\ &= - 2.1 \times {10^{ - 18}}\;{\rm{J}}\end{aligned}\)

Evaluation of the final electric potential energy

The final potential energy can be calculated as:

\(\begin{aligned}P{E_2} &= \frac{{k{q_1}{q_2}}}{{{r_2}}}\\ &= \frac{{\left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right)\left( { - 1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}}{{\left( {0.100\;{\rm{nm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{nm}}}}} \right)}}\\ &= - 2.3 \times {10^{ - 18}}\;{\rm{J}}\end{aligned}\)

Evaluation of the difference in the potential energies

The difference in the potential energies or the loss in the potential energy can be calculated as:

\(\begin{aligned}\Delta PE &= P{E_1} - P{E_2}\\ &= - \left( {2.1 \times {{10}^{ - 18}}\;{\rm{J}}} \right) - \left( { - 2.3 \times {{10}^{ - 18}}\;{\rm{J}}} \right)\\ &= \left( {0.2 \times {{10}^{ - 18}}\;{\rm{J}}} \right)\left( {\frac{{1\;{\rm{eV}}}}{{1.6 \times {{10}^{ - 19}}\;{\rm{J}}}}} \right)\\ &= 1.25\;{\rm{eV}}\end{aligned}\)

Thus, the electric potential energy lost during the reaction is \(1.25\;{\rm{eV}}\).