1.An exothermic reaction at standard conditions (298 K and 1 atm) has q c = 5.00

ID: 973736 • Letter: 1

Question

1.An exothermic reaction at standard conditions (298 K and 1 atm) has

qc = 5.00 kJ/mole (heat released) and DV = + 4.95 × 10-2 m3/mole (volume increases).

Use w = PDV to determine the work performed in both J and kJ by the gas.

Use 1 atm = 1.01 × 105 J/m3 in your equation.

This works because 1 J/m3 = 1 kg/(m·s2) = 1 Pa. See Table 5.2.

Use significant figures. Show all signs, units, and conversion factors.

Use DU = q + w to determine the change in internal energy in both J and kJ.

Use significant figures. Show all signs, units, and conversion factors

Use DH = DU + PDV to determine the change in enthalpy in both J and kJ.

Use significant figures. Show all signs, units, and conversion factors

LEt's calculate the Work:

W = -1.01x105 J/m3 * 4.95x10-2 m3/mol
W = -4999.5 J/mol or simple -4.9995 kJ/mol

dU = -5 kJ/mol - 4.9995 = -9.9995 kJ/mol or -10 kJ/mol. In J is -9999.5 J/mol or 10000 J/mol

Finally, dH:
dH = -10000 + (-5000)
dH = -15000 J/mol or 15 kJ/mol

Hope this helps

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