Question

Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.135 nm , and their final separation is 0.105 nm .

How much electric potential energy was lost in this reaction (in units of eV)?

Answer 1

Let the equation for initial potential energy be U_i = -k*q_p*q_e/r_i
Let the equation for final potential energy be U_f = -k*q_p*q_e/r_f

k is Columb's constant
q_p = charge of proton
q_e = charge of electron
r_i = initial distance
r_f = final distance

We are trying to find the difference U_f - U_i

U_f - U_i = [-k*q_p*q_e/r_f] - [-k*q_p*q_e/r_i]

We can pull out -k, q_p, and q_e to simplify the equation down a bit

U_f - U_i = -k*q_p*q_e*[(1/r_f) - (1/r_i)]

q_p and q_e equal the same thing except opposite signs. q_p=+e and q_e= -e where e=1.602*10^(-19) C.

So plugging q_p = +e and q_e=-e we have

U_f - U_i = +k*e^2*[(1/r_f) - (1/r_i)]

Now plugging everything in, we have k=8.99*10^9, e=1.602*10^(-19) C, r_f= 0.105*10^(-9) m, r_i = 0.135*10^(-9) m (I converted the distances from nm to m)

We get that

U_f - U_i = 4.88 *10^(-19) J

Convert to electronvolts

4.88*10^(-19) J *(6.2415*10^18 eV/J) = 3.05 eV.

Hope this helps you.

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