# Homework Help: Chemical reaction

1. Jan 13, 2015

### geoffrey159

1. The problem statement, all variables and given/known data
A simple and very violent chemical reaction is H + H -> H_2 + 5 eV (1eV = 1.6x10^-19 J, a healthy amount of energy on the atomic scale). However, when hydrogen atoms collide in free space they simply bounce apart! The reason is that it is impossible to satisfy the laws of conservation which releases energy. Can you prove this ? You might start by writing the statements of conservation of momentum and energy. Be sure to include the energy of reaction in the energy equation, and get the sign right. By eliminating the final momentum of the molecule from the pair of equations, you should be able to show that the initial momenta would have to satisfy an impossible condition.

2. Relevant equations
conservation of momentum and energy

3. The attempt at a solution

I assume that two colliding atoms of hydrogen with speed $v_1$ and $v_2$ do make a molecule of dihydrogen moving at speed $v$, and want to prove that it leads to a contradiction.

Momentum before and after collision is conserved, so
$\left. \begin{array}{} |\vec P_i|^2= m^2 ( v_1^2 + v_2^2 ) \\ |\vec P_f| ^2 = 4 m^2 v^2 \\ \vec P_i = \vec P_f \end{array} \right\} \Rightarrow v_1^2 + v_2^2 =4 v^2$

By conservation of total energy, and because 5eV of energy is realeased in the chemical reaction,
$K_i = K_f - 5eV \Rightarrow \frac{1}{2} m (v_1^2+v_2^2) = \frac{1}{2}m v^2 - 5eV \Rightarrow \frac{3}{2} m v^2 = -5eV \Rightarrow v^2 < 0$

This is impossible so the assumption is wrong and two colliding hydrogen atoms do not make a molecule of dihydrogen and just bounce apart.

Does it look right to you ?

2. Jan 13, 2015

### BvU

Momentum2 before is not sum of v squareds but (sum of v) squared

3. Jan 13, 2015

### geoffrey159

Sorry for the error, I was too much in a hurry

By momentum conservation,

$\left. \begin{array}{} |\vec{P_i}| = m |v_1-v_2| \\ |\vec{P_f}| = 2mv \\ \vec P_i = \vec P_f \end{array} \right\} \Rightarrow |\vec P_i|^2 = |\vec P_f| ^2 \Rightarrow v^2 = \frac{1}{4} (v_1^2 + v_2^2 - 2 v_1v_2)$

By total energy conservation,

$K_i = K_f - 5eV \Rightarrow \frac{1}{2} m (v_1^2+v_2^2) = m v^2 - 5eV \Rightarrow \frac{1}{4} m (v_1^2 + v_2^2) + \frac{1}{2}mv_1v_2 = -5eV$

But $0\le v_1^2 + v_2^2$, and $0\le v_1v_2$ because $v_1$ and $v_2$ are magnitudes.

In the end, I get that

$0 > -5eV = \frac{1}{4} m (v_1^2 + v_2^2) + \frac{1}{2}mv_1v_2 \ge 0$, which is a contradiction.

Are you ok with that now ?

4. Jan 13, 2015

### BvU

No, $v_1$ and $v_2$ are vectors.

5. Jan 13, 2015

### geoffrey159

They are magnitudes here. What is wrong according to you ?

6. Jan 13, 2015

### BvU

$\vec P_i = m(\vec v_1 + \vec v_2) \ \ \Rightarrow |\vec P_i |^2 = v_1^2 + v_2^2 + 2 \vec v_1 \cdot \vec v_2$
Your minus sign tells me you only consider head-on collisions, which may or may not be OK -- if you explain

7. Jan 13, 2015

### geoffrey159

It makes sense, thanks !

Fortunately, it does not change much the calulations and

$0> -5eV = \frac{1}{4} m (v_1^2 + v_2^2) - \frac{1}{2}m\vec v_1.\vec v_2 = \frac{1}{4} m |\vec v_1 - \vec v_2| ^2 \ge 0$

Contradiction. Is that Ok now ?

8. Jan 13, 2015

### BvU

Looks good.

MIght have avoided this fuzz by looking in the center of mass system, where |P| = 0

9. Jan 13, 2015

### geoffrey159

Thank you for your help !

10. Jan 13, 2015

### BvU

You're welcome