Answer to Question #134042 in Organic Chemistry for Michele Valier

Question

Calculate the reaction free energy of

H2(g) + I2(g) 2HI(g)

when the concentrations are 0.006 mol L-1 (H2), 0.03 mol L-1 (I2), and 1.27 mol L-1 (HI), and the temperature is 700K. For this reaction Kc = 54 at 700 K.

+52,100 kJ mol-1

+52 kJ mol-1

-29.8 kJ mol-1

+29.8 kJ mol-1

Answer 1

$\Delta$$\mathring{G}$ = RTl_nK_c

R = 8.314 j/kmol

T = 700 K

K_c = 54

$\frac{[HI][HI]}{[H2][I2]} =$ $\frac{1.27][1.27]}{[0.006][0.03]}$

$=$ 8960.55

$\Delta G =\Delta$$\mathring{G}$ + RTl_n$\theta$

$= -RTl$_n(K_c) $+ RTl$_n$\theta$

$= RTl$_n$\frac{\theta}{Kc}$

$= 8.314 \times 700 \times l\frac{8960.55}{54}$

$= 29.8 kj/mol$