Some thermo and gas problem questions?

Question

Here are a couple of more questions I have after reviewing:

For which of the following equations is the enthalpy change at 25degreesC and 1 atm equal to standard deltaHformation of

HCOOH(l)?

a. CO2(g) + H2(g) > HCOOH(l)

b. CO(g) + H2O(l) . HCOOH (l)

c. C(s) + 2H(g) + 2O(g) > HCOOH (l)

d. C(g) + 2H(g) + 2O(g) > HCOOH (l)

e. C(s) + H2(g) + O2(g) > HCOOH (l)

ANSWER IS E, BUT WHY?

Next:

THe molecular weight of an unknown gas was measured by an effusion experiment. It was found that it took 64 seconds for the gas to effuse, whereas nitrogen required 48 seconds. The molecular mass of the gas is

a. 16 g/mol

b. 24 g/mol

c. 30 g/mol

d. 40 g/mol

e. 50 g/mol

ANSWER IS A, BUT WHY AGAIN?

thanks

Answer 1

For the first one, the definition of standard enthalpy of formation that I always remember is, the amount of energy required to produce one mole of something in its natural state at S.T.P from its constituent elements in their natural state.  Now, the first one is clear, they want a the equation that represents the formation of HCOOH to represent the standard enthalpy of formation.  So we need the constituent elements, C, H, and O, in their natural states C (s), H2 (g), and O2 (g), to form the product HCOOH, in it's natural state, HCOOH (l)--So Letter "E" is the answer.

This one is simply Grahm's law of effusion:

Rate 1/Rate 2 = Sqrt(M2/M1); where M = Molecular Weight and rates are proportional to times given above, and 1 will be the unknown gas.  Also, the molecular weight of nirtogen gas is 28 g/mol, so:

64 sec / 48 sec = Sqrt ((28 g/mol )/ X ):

X = 15.75 g/mol = ~16 g/mol