Question

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

Answer 1

- 431.15 kJ/mol.

Step-by-step 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.