A sample consisting of 22.7 g of a nongaseous, unstable com- pound X is placed i

Question

A sample consisting of 22.7 g of a nongaseous, unstable com- pound X is placed inside a metal cylinder with a radius of

8.00 cm, and a piston is carefully placed on the surface of the compound so that, for all practical purposes, the distance be- tween the bottom of the cylinder and the piston is zero. (A hole in the piston allows trapped air to escape as the piston is placed on the compound; then this hole is plugged so that nothing inside the cylinder can escape.) The piston-and-cylinder appara- tus is carefully placed in 10.00 kg water at 25.00 C. The baro- metric pressure is 778 torr.

When the compound spontaneously decomposes, the piston moves up, the temperature of the water reaches a maximum of 29.52 C, and then it gradually decreases as the water loses heat to the surrounding air. The distance between the piston and the bottom of the cylinder, at the maximum temperature, is 59.8 cm. Chemical analysis shows that the cylinder contains 0.300 mol carbon dioxide, 0.250 mol liquid water, 0.025 mol oxygen gas, and an undetermined amount of a gaseous element A.

It is known that the enthalpy change for the decomposition of X, according to the reaction described above, is 1893 kJ/mol X. The standard enthalpies of formation for gaseous carbon dioxide and liquid water are 393.5 kJ/mol and 286 kJ/mol, respectively. The heat capacity for water is 4.184 J/ C g. The conversion factor between L atm and J can be determined from the two values for the gas constant R, namely, 0.08206 L atm/K mol and 8.3145 J/K mol. The vapor pressure of water at 29.5 C is 31 torr. Assume that the heat capacity of the piston- and-cylinder apparatus is negligible and that the piston has negligible mass.
Given the preceding information, determine

The formula for X.

The pressure–volume work (in kJ) for the decomposition of

the 22.7-g sample of X.

Themolarchangeininternalenergyforthedecompositionof

X and the approximate standard enthalpy of formation for X.

Explanation / Answer

The formula for X.

Chemical analysis shows that the cylinder contains

0.300 mol carbon dioxide,

Co2----- produced from ---0.300 mol carbon and 0.600 mol oxygen

==============================

0.250 mol liquid water,

H2O

H2O is produced from 0.500 mol H and 0.250 mol O.

==============================

0.025 mol oxygen gas,

O2 i is produced from 0.050 mol O.

==============================

c

0.300 mol

H

0.500 mol

O

0.600 + 0.250 + 0.050 = 0.900 mol

HENCE formula for X.---------C 0.3 H 0.5 O 0.9

C 0.3 H 0.5 O 0.9

Multiplying by 10 ABOVE EQUATION

C3H5O9 ------------------answer

c

0.300 mol

H

0.500 mol

O

0.600 + 0.250 + 0.050 = 0.900 mol

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