Calculate the work done when 36.0 g of tin dissolves in excess acid at 1.00 atm and 30.1 C.

Answer 1

Each mole of solid Sn that reacts produces 1 mole of N2 gas. The difference in the volume occupied by the products and reactants will be dominated by the volume of the gas produced in the reaction, and we will ignore any change in the volume of the solid and liquid phases in the system

For a system undergoing a change in volume at constant pressure, the work done *by* the system *on* the environment is given by:

work = P *(final volume - initial volume)

In this case, the difference between the final and initial volumes of the system is essentially equal to the volume of the H2 gas produced by the reaction.

Assuming ideal gas behavior, 1 mole of H2 gas at 30.1 C (303.25 K) and 1 atm (1.013*10^5 pascals) occupies a volume of:

V = n*R*T/P

so the work done per mole of Sn that reacts is:

work/mole = 1mol*R*T

work/mole = 1mol * 8.314J/mol*K * 303.25K

work/mole = you do it.
The atomic mass of Sn is 118.71 So a mass of 36 grams is .303 moles.

The work done by the reaction of 36 gm of Sn under the specified conditions is therefore:

0.303 * work/mole = ....