Answer to Question #319837 in Physical Chemistry for HRMSCA

Question

Calculate the differences in energy between ΔU and ΔH for the following reactions at constant pressure at 25oC.( use ΔH = ΔU + PΔV)

a) CH3CH2CH3(g) + 5O2(g) ⇔ 3CO2(g) + 4H2O (l)

b) C6H12O6 (s)+6O2(g) ⇔ 6 CO2(g) + 6 H2O(l)

c) Mg (s) + 2H + (aq) ⇔ Mg2(aq) + H2(g)

Answer 1

ΔH = ΔU + P * ΔV

ΔH = ΔU + n * R * T;

(ΔU - ΔH) = - n * R * T;

T = 25 C = 298 K;

R = 8.314 J*K^-1mol^-1;

(ΔU - ΔH) = -n * 8.314 * 298 = -n * 2477.6;

If n = 1 mol, then: (ΔU - ΔH) = -2477.6 J = -2.478 kJ.

If n = 2 mol, then: (ΔU - ΔH) = -2 * 2477.6 = -4955.1 J = -4.955 kJ.

If n = 3 mol, then: (ΔU - ΔH) = -3 * 2477.6 = -7432.8 J = -7.433 kJ.

If n = 4 mol, then: (ΔU - ΔH) = -4 * 2477.6 = -9910.4 J = -9.910 kJ.

If n = 5 mol, then: (ΔU - ΔH) = -5 * 2477.6 = -12388 J = -12.388 kJ.