When 25.0 mL of 1.00 M NaOH(aq) at 25.0 degree C is mixed with 25.0 mL of 1.00 M

Question

When 25.0 mL of 1.00 M NaOH(aq) at 25.0 degree C is mixed with 25.0 mL of 1.00 M HCl (aq) at 26.5 degree C, the temperature of the resulting solution rises to 32.5 degree C. Assume S_ = 4.087 J/g degree C. Is this reaction endothermic or exothermic? Calculate the mass of the solution. Assume d_solution = 1.04 g/mL. Calculate Delta T of the reaction. You will need to average the two initial temperatures to get the initial temperature of the solution. Calculate using Equation 8. Calculate Delta H for the reaction. (Delta H_ = -q_solution) Calculate the moles of water formed by the reaction. Remember that: HCl (aq) + NaOH (aq) rightarrow H_2 O(1) + NaCl (aq) Finally, calculate Delta H for the reaction per mode of water. That is, divide your answer to part (e) by your answer to part (f).

Explanation / Answer

1.since there is a rise in temperature during the course of reaction, the reaction is exothermic.

2. Volume of mixture =25+25=50ml, density of solution =1.04 g/ml, mass of solution = 50*1.05=52 gm

3. Average initial temperature = (25+26.5)/2= 25.75 deg.c. This is the initial average temperature. Final temperature = 32.5 deg.c. Rise in temperature= 32.5-25.75= 6.75 deg.c

4. q = mcp*temperature difference= 52*4.087*6.75 =1434 joules

5. deltaH=-q=-1434

6. moles of NaOH= molarity* Volume(L)= 1*25/1000 =0.025

enthalpy change= -1434/0.025 j/mole=

-57360 J/mole=-57.360 Kj/mole

7. from the reaction, mole of water formed= moles of NaOH used= 0.025

hence enthalpy change/mole of water= -57.360 Kj/mole

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