Question

50.0mL of .500M NaOH is added to 25.0mL of .500M HCL in a calorimeter BOth are at an initial temperature of 25 C ans final temperature is 27.21 C. Assume the mass of the final solution is 76.0 g and the Cs of the solution is 4.18 J/gC. what is the value of q for water, what is the value of q for the reaction, what is the change of H of the reaction in kj/mol?

Answer 1

q = ms ΔT = = 76 x 4.18 x(27.21-25) = 702.07 joules

   NaOH    +     HCl -------->    H2O   + NaCl

t=0      25                     12.5                                       millimoles

t=t      12.5                   0                 12.5      12.5 millimoles

for   12.5 millimoles    --------->    energy   is 702.07 joules

for      1 millimoles     ---------->    702.07/12.5 Joules

for 1 mole= 1000millimoles ------->   1000x (702.07/12.5) =56165.6 joules

ΔH = 56.165.6KJ /moles