Question

Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 30.0°C. Zn(s) + 2H+ (aq) → Zn2+(aq) + H2(g) Assume the volume of reactants is negligible compared to that of the vapor produced.

Answer 1

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)×$\frac{1molZn}{65,38g}$×$\frac{1molH_{2}(g)}{1molZn}$ = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = 190,1L

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL×$\frac{101,325}{1atmL}$ = 19,26 kJ

I hope it helps!