When 100.0 cm-3of 1.00 mol dm-3 NaOH is added to 100.0 ml of 1.00 moldm-3 HCl in aninsulated container the temperature rises from 21.0°C to 34.6°C.NaOH(aq) + HCl (aq) → NaCl(aq) + H2O(l) Calculate the enthalpy change for theneutralization reaction. State any assumptions made
Question
When 100.0 cm-3of 1.00 mol dm-3 NaOH is added to 100.0 ml of 1.00 moldm-3 HCl in aninsulated container the temperature rises from 21.0°C to 34.6°C.NaOH(aq) + HCl (aq) → NaCl(aq) + H2O(l) Calculate the enthalpy change for theneutralization reaction. State any assumptions made
Solution
To calculate the enthalpy change for the neutralization reaction, we need to follow these steps:
- First, we need to calculate the heat absorbed by the solution. We can use the formula q = mcΔT, where m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
Assuming the density of the solution is approximately the same as water (1 g/cm³), the mass of the solution is 200.0 g (100.0 cm³ of NaOH + 100.0 cm³ of HCl). The specific heat capacity of water is 4.18 J/g°C. The change in temperature is 34.6°C - 21.0°C = 13.6°C.
So, q = (200.0 g)(4.18 J/g°C)(13.6°C) = 11344 J.
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Next, we need to calculate the number of moles of NaOH and HCl that reacted. Since the concentration of each is 1.00 mol/dm³ and the volume of each is 100.0 cm³ (or 0.1 dm³), the number of moles of each is (1.00 mol/dm³)(0.1 dm³) = 0.1 mol.
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Finally, we can calculate the enthalpy change (ΔH) for the reaction. ΔH is equal to the heat absorbed by the solution divided by the number of moles of NaOH (or HCl) that reacted, and then multiplied by -1 (since the reaction is exothermic and the heat is released, not absorbed).
So, ΔH = -q/n = -(11344 J) / (0.1 mol) = -113440 J/mol = -113.44 kJ/mol.
Assumptions made:
- The density of the solution is the same as water.
- The specific heat capacity of the solution is the same as water.
- The reaction is perfectly efficient (all the heat released by the reaction is absorbed by the solution).
- The reaction goes to completion (all the NaOH and HCl react).
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