Answer :
Energy is released (in kj) in the fusion reaction of 2h to yield 1 mol of 3he is 3.204 × 10⁸ kJ / mol.
Calculation:-
2 ²₁ H -----> ³₂He + ¹₀n
mass 2.041 3.0160 1.008664
mass of product = 2 × 2.0141 = 4. 0282
mass of reactant = 3.0160 + 0.08664 = 4.02464
mass defect = mass of product - mass of reactant
= 4.0282 - 4. 2464
= 3.56 * 10⁻³ amu
= 3.56 * 10⁻³ amu × 1.66 × 10 ⁻²⁷ kg
= 5.9115 × 10⁻³⁰ kg
energy released = mc²
= 5.9115 × 10⁻³⁰ kg × (3 × 10⁸)²
= 5.32 * 10 ⁻¹³ J
conversion:-
5.32 * 10 ⁻¹³ J * 6.022 × 10²³ / 1 mol × 1 kj/10³ j
= 3.204 × 10⁸ kJ / mol
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