When .03 mol Na is added to 100g of water, the temp. of the resulting solution rises from 25C to 37.9C. If the specific heat of the solution is 4.18 J/(g C), calculate delta H for the reaction, 2Na(s) + 2H2O(l) --> 2NaOH(aq) + H2(g)
q = mass x specific heat x delta T and that is for 0.03 mol Na or 0.69 grams Na. You want q for the reaction which is for 46 g Na. You may need to tweak the 0.69 and 46 since I didn't look up the atomic mas Na but those numbers are close.
To calculate ΔH for the reaction, we need to use the formula:
ΔH = q / n
Where:
- ΔH is the enthalpy change (in Joules) for the reaction
- q is the heat absorbed or released by the reaction (in Joules)
- n is the amount of substance involved in the reaction (in moles)
First, we need to calculate the heat absorbed or released by the reaction (q).
q = mcΔT
Where:
- q is the heat absorbed or released by the reaction (in Joules)
- m is the mass of the solution (in grams)
- c is the specific heat of the solution (in J/(g°C))
- ΔT is the change in temperature (in °C)
Given:
- The mass of water (m) = 100g
- The change in temperature (ΔT) = 37.9°C - 25°C = 12.9°C
- The specific heat of the solution (c) = 4.18 J/(g°C)
Calculating q:
q = (100g) * (4.18 J/(g°C)) * (12.9°C)
q = 5403 J
Next, we need to calculate the amount of substance (moles) involved in the reaction (n).
Given:
- 0.03 mol Na is added
So, n = 0.03 mol
Finally, we can calculate ΔH using the formula:
ΔH = q / n
ΔH = 5403 J / 0.03 mol
ΔH ≈ 180100 J/mol
Therefore, the enthalpy change (ΔH) for the reaction 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) is approximately 180100 J/mol.