Question

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

Answer #1

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

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