Answer:
–3.47 KJ
Explanation:
We'll begin by calculating the number of mole in 3.60 g of NaHSO₄. This can be obtained as follow:
Mass of NaHSO₄ = 3.60 g
Molar mass of NaHSO₄ = 23 + 1 + 32 + (16×4)
= 23 + 1 + 32 + 64
= 120 g
Mole of NaHSO₄ =?
Mole = mass / molar mass
Mole of NaHSO₄ = 3.6 / 120
Mole of NaHSO₄ = 0.03 mole
Next, the balanced equation for the reaction. This is illustrated below:
2NaHSO₄ —> Na₂SO₄ + H₂O + SO₃
ΔH = –231.3 KJ
From the balanced equation above,
2 moles of NaHSO₄ reacted to produce –231.3 KJ of heat energy.
Finally, we shall determine the enthalpy change involved when 3.6 g (i.e 0.03 mole) of NaHSO₄ reacted. This can be obtained as follow:
From the balanced equation above,
2 moles of NaHSO₄ reacted to produce –231.3 KJof heat energy.
Therefore, 0.03 mole of NaHSO₄ will react to produce = (0.03 × –231.3) / 2
= –3.47 KJ of heat energy.
Thus, the enthalpy change (ΔH) involved in the reaction is –3.47 KJ.
