What is the value of w for each of the following processes at 298K and 1atm? Hin

Question

What is the value of w for each of the following processes at 298K and 1atm?

Hint 1: you may wish to write balanced equations for these processes

Hint 2: since you are not starting out with one mole in these processes, any (delta)n(gas) you get from the balanced equation will need to be scaled appropriately.

A) condensation of 11.4g of water(answer in joules)

B) reaction of 5.00g Na(solid) and excess Cl2(gas) to produce NaCl(solid)

-hint the change in the amount of Cl2 is dictated by the amount of Na in the reaction (answer in joules)

C) decomposition of 12.5g KClO3 into KCl(solid) and O2(gas). (Answer in joules)

Explanation / Answer

similar one for u

work = P x deltaV
using pv = nrt to determine the volume of H2O(g)
v = nrt/p
v = (11.1/18)(0.0821L-atm/mole-K)(298K) / 1atm = 15.087L
so, 15.087L H2O vapor is condensing to 11.1ml
work = 1atm x 15.076L x 101.325J/L-atm = 1527.57J

2Na + Cl2 --> 2NaCl
5.4g Na / 23g/mole = 0.235moles Na, this requires 0.1175moles Cl2
v = nrt/p = 0.1175moles Cl2 x 0.0821L-atm/mole-K x 298K / 1atm = 2.87L Cl2 gas
w = P x deltaV = 1atm x 2.87L x 101.325J/L-atm = 290.8J

10.1g KClO3 / 122.45g/mole = 0.082moles KClO3
2KClO3 --> 2KCl + 3O2
v = nrt / p = 0.082moles x 0.0821L-atm/mole-K x 298K / 1atm = 2L KClO3 gas
2L KClO3 gas produces 3L O2 gas
w = P x deltaV = 1atm x 1L x 101.325J/L-atm = 101.325J

6.3g H2 / 2g/mole = 3.15moles H2
H2 + CO2 --> H2O + CO, balanced
volume doesn't change, work = 0

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