When 2.50 g of Ba(s)is added to 100.00 g of water in a container open to the atmosphere,the reaction shown below occurs and the temperature of the resulting solution rises from 22.00 °C to 40.32 °C.If the specific heat of the solution is 4.18 J $g ^ { - 1 }$ ${ } ^ { \circ } \mathrm { C } ^ { - 1 }$ ,calculate $\Delta H$ for the reaction,as written. Ba(s)+ 2H₂O(l)? Ba(OH)₂(aq)+ H₂(g) $\Delta _ { \mathrm { r } } H$ = ?

Question

Quizplus · Accepted Answer

The heat absorbed by the solution can be calculated using q = mcΔT, where m is the mass of the solution (102.50 g), c is the specific heat (4.18 J/g°C), and ΔT is the temperature change (18.32°C). This gives q = 102.50 g * 4.18 J/g°C * 18.32°C = 7850.67 J or 7.85 kJ. Since the reaction releases heat, ΔH is negative. The moles of Ba is 2.50 g / 137.33 g/mol = 0.0182 mol. ΔH per mole is -7.85 kJ / 0.0182 mol = -431 kJ/mol.

When 250 G of Ba(s)is Added to 100 g−1g ^ { - 1 }g−1

When 250 G of Ba(s)is Added to 100 $g ^ { - 1 }$