Question

Ist attempt See Periodic Table O See Hi Calculate the final temperature of 100.0 mL of water (d=1.00 g/mL) at 20.2 °C 1% = 75.3 J/mol·°C)] after adding 100 mL of concentrated HNO3 (14.7 M, AH soln --33.3 kJ/mol).

Answer 1

When a strong acid is added to water an exothermic reaction takes place and this reaction causes the increase in temperature of the water.

Assuming density does not change with mixing, we first find the heat of reaction. Also total volume after mixing = 110 ml

Q_{​​​​​​rxn} = ∆H * molarity * volume = -33.3 * 14.7 * 0.01 KJ = 4.8951 KJ = - 4895.1 J

This cause a change in temperature and this change is related as Q_{​​​​​sol} = No of moles of water * C_{​​​​​​p} in J/mol°C * ∆T = - Q_{​​​​​​rxn}

4895.1 = (110ml * 1g/ml / 18g) moles * 75.3 J/mol°C * ∆T

∆T = 4895.1 / (6.11 * 75.3) = 10.64 °C

T_{​​​​​​f} - T_{​​​​​​i} = 10.64

=> T_{​​​​​​f} = 10.64 + T_{​​​​​​i} = 10.64 + 20.2 = 30.84°C

So final temperature of the mixture is 30.84°C