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a student performed the procedure described in the lab manual using 49.9 ml of 1.0 m hci and 49.3 ml of 1.0 m naoh. reaction caused the temperature inside the calorimeter to increase from 17.8 oc to 25.5 oc. what is the value of the enthalpy change for this reaction (haqueous) in units of kj/mol?

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Final answer:

To calculate the enthalpy change for this reaction, use the equation q = m * c * ∆T. The enthalpy change per mole of HCl is 64.4 kJ/mol.

Step-by-step explanation:

To calculate the enthalpy change for this reaction, we can use the equation:

q = m * c * ∆T

where q is the heat released or absorbed, m is the mass of the solution, c is the specific heat capacity of the solution, and ∆T is the change in temperature. In this case, we have a total volume of 99.2 mL, which corresponds to a mass of 99.2 g (assuming a density of 1.00 g/mL). The change in temperature is 7.7 °C (25.5 °C - 17.8 °C). Assuming the specific heat capacity of the solution is the same as that of water (4.18 J/g·°C), we can calculate the heat produced by the reaction:

q = 99.2 g * 4.18 J/g·°C * 7.7 °C = 3221.85 J

To convert joules to kilojoules, we divide by 1000:

3221.85 J / 1000 = 3.22 kJ

Now we need to convert this value to the enthalpy change per mole of HCl:

Since the reaction uses equal moles of HCl and NaOH, the enthalpy change for the reaction is the same as the enthalpy change for 1 mole of HCl. Therefore, the enthalpy change is:

ΔH = 3.22 kJ / 0.0500 mol = 64.4 kJ/mol

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