Answer to Question #82008 in General Chemistry for Manvir Nahal

Question

A coffee-cup (constant pressure) calorimeter is used to carry out the following reaction in 82.6 mL water (where X is a hypothetical metal):

X + 2 H2O → X(OH)2 + H2

In this process, the water temperature rose from 25.0 °C to 37.5 °C. If 0.00610 mol of "X" was consumed during the reaction, what is ΔH of this reaction with respect to the system in kJ mol-1 ?

The specific heat of water is 4.184 J g-1 °C-1

Answer 1

Heat absorbed by calorimeter

q = CmΔT

q = 4.184J/g·°C*82.6g*(37.5°C - 25°C)

q = +4.32 kJ absorbed by calorimeter so therefore q= -4.32 kJ for heat released from the reaction

and per mole

Heat evolved = ( -4.32 kJ / 0.0061 mol of metal)

ΔH = - 708.2 kJ/mol