When 1.50g of Ba is added to 100g of water in a container open tothe atmosphere, the reaction shown below occurs and the temperatureof the resulting solution rises from 22 degrees to 33.10 degrees. If the specific heat of the solution is 4.18J / (g*C), calculatedelta H for the reaction, as written. Ba (s) + 2H2O (l) yields Ba (OH) 2 (aq) + H2

- 431.15 kJ/mol.

Explanation:

Firstly, we can calculate the amount of heat (Q) released by the solution using the relation:

Q = m. c.ΔT,

where, Q is the amount of heat released from the solution (Q = ? J).

m is the mass of solution (m = 1.5 g + 100 g = 101.5 g).

c is the specific heat capacity of solution (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = 33.1°C - 22°C = 11.1°C).

∴ Q = m. c.ΔT = (101.5 g) (4.18 J/g.°C) (11.1°C) = 4709.4 J.

To find ΔH:

∵ ΔH = Q/n

no. of moles of Ba (n) = mass/atomic mass = (1.50 g) / (137.3270 g/mol) = 0.011 mol.

∴ ΔH = - Q/n = (4709.4 J) / (0.011 mol) = - 431.15 kJ/mol.

The negative sign is not from calculation, but it is an indication that the reaction is exothermic.