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

Question

Question

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

q_c = − 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

Answer 1

Answer #1

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

Answer 2

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