A 2.09 g piece of zinc metal is added to a calorimeter containing 1.50 L of 0.600 M HCl. This results in a reaction where zinc chloride and hydrogen gas is formed, and the temperature of the solution increases by 0.780°C. The density of the solution is 1.00 g/mL, and its specific heat capacity is the same as water, what is the enthalpy change, per mole of zinc, for the reaction?
(a) 65.8 kJ
(b) –2.34 kJ
(c) –4.90 kJ
(d) –153 kJ
(e) –136 kJ
Amount of heat leberated , q = mcdt
Where
m = mass of solution = volume x density
= 1.50 L x(1000 mL/L) x 1.00(g/mL)
= 1500 g
c = specific heat capacity of solution = 4.184 J/goC
dt = raise in temperature = 0.780 oC
Plug the values we get Q = 4895.3 J
This is the amount of heat leberated by 2.09 g of Zn
So 2.09 g of Zn leberates 4895.3 J of heat
1 mole = 65.4 g of Zn leberates (65.4x4895.3)/2.09 = 153x103 J
= 153 kJ
Therefore enthalpy change, per mole of zinc is -153 kJ
So option (d) is correct
Get Answers For Free
Most questions answered within 1 hours.