# Question 8a2b4

Dec 15, 2015

$4.750089 \cdot {10}^{11} \text{J/g}$

#### Explanation:

The idea here is that you need to use Einstein's equation to convert the difference in mass between the helium-4 atom and the sum of the hydrogen-3 and hydrogen-1 atoms to energy.

$\textcolor{b l u e}{E = m \cdot {c}^{2}} \text{ }$, where

$E$ - energy
$m$ - mass, expressed in kilograms
$c$ - the speed of light in vacuum, equal to ${\text{299,792,458 m s}}^{- 1}$

The idea here is that the atomic mass of the helium-4 atom will be smaller than the sum of the atomic masses of the hydrogen-3 and hydrogen-1 atoms because some of that mass is converted to energy.

The sum of the hydrogen-3 and hydrogen-1 atoms will be

${m}_{\text{total" = "3.016049 u" + "1.007825 u" = "4.023874 u}}$

The difference between this value and the actual mass of the helium-4 atom, also called mass defect, will be

${m}_{\text{diff" = "4.023874 u" - "4.002603 u" = "0.021271 u}}$

Now, the unified atomic mass unit, $u$, which is defined as $\frac{1}{12} \text{th}$ of the mass of an unbound neutral carbon-12 atom, is equivalent to

$\text{1 u" = 1.660539 * 10^(-27)"kg}$

Use this conversion factor to convert the mass defect from unified atomic mass units to kilograms

0.021271 color(red)(cancel(color(black)("u"))) * (1.660539 * 10^(-27)"kg")/(1color(red)(cancel(color(black)("u")))) = 3.532133 * 10^(-29)"kg"

Use Einstein's equation to calculate the energy released per atom of hydrogen-3 and hydrogen-1

E = 3.532133 * 10^(-29) * ("299,792,458")^2 overbrace("kg m"^2"s"^(-2))^(color(blue)("Joules"))

$E = 3.174523 \cdot {10}^{- 12} \text{J}$

Now, to get the energy released per gram of reactant, use Avogadro's number to first convert the energy from Joules per atom to Joules per mole

$3.174523 \cdot {10}^{12} \text{J"/color(red)(cancel(color(black)("atom"))) * (6.022 * 10^(23)color(red)(cancel(color(black)("atoms"))))/"1 mole" = 1.911698 * 10^12"J/mol}$

Now, the total atomic mass of the reactants is equal to $\text{4.023874 u}$, use the fact that

$\text{1 u " = " 1 g/mol}$

to get a total molar mass of the reactants of $\text{4.023874 g/mol}$. This means that the energy released per gram of reactant will be

1.911698 * 10^(12)"J"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mol"))))/"4.023874 g" = color(green)(4.750089 * 10^(11)"J/g")#

The answer is rounded to seven sig figs, the number of sig figs you have for the given atomic masses.