Question

When 60.0 mL of a 0.400 M solution of HNO3 is combined with 60.0 mL of a 0.400 M solution of NaOH in the calorimeter described in question #1, the final temperature of the solution is measured to be 26.6 ˚C. The initial temperature of the solutions is 24.0 ˚C. Assuming the specific heat capacity of the final solution is 3.90 J·g–1 ·˚C–1 and the density of the final solution is 1.04 g/mL, calculate qrxn. Hint: start with qrxn + qsoln + qcal = 0. HNO3(aq) + NaOH(aq) → H2O(l) + NaNO3(aq)

Answer 1

I am assuming calorimeter doesn't absorb any heat.
qrxn + qsoln + qcal = 0 becomes,
qrxn + qsoln = 0
qrxn = -qsoln

qsoln = m*C*delta T
m = density * volume of solution
    = 1.04 g/mL * (60 + 60) mL
    = 124.8 g

qsoln = m*C*delta T
   = 124.8 * 3.9* ( 26.6 - 24)
   =1265.5 J

qrxn = -qsoln
    = -1265.5 J

Number of moles of acid or base reacted are equal
Number of moles, n = M * V = 0.4 M * 0.06 L = 0.024 mol

In J/mol, qrxn can be written as,
qrxn = - 1265.5 J / 0.024 mol
          = -52729 J/mol
          = -52.729 KJ/mol
Answer: -52.729 KJ/mol