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

Question

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

Explanation / Answer

a) this reaction is an exothermic reaction, because heat is given out by the system during course of reaction.

b) Mass of the solution = ( volume of NaOH + volume of HCl)* density of solution

=(25+25)*1.04= 53g

c) mean of initial temperature = (25+26.5)/2= 25.75oC

delta of T= Tf -Ti = 32.5-25.75= 6.75o C

d) q=2.303nRTlog(Vf/Vi)=2.303*1*32.5*log(50/2)= 81.34J/mol

e) delta H= -q = -81.34 J/mol

f) HCl + NaOH > H2O + NaCl

from the reaction it is clear that 1mole of HCl and 1mole of NaOH gives you 1mole pf H2O

g) delta H for the reaction per mole of water= delta H/ moles of water= -81.34 J/mol

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