a) First, to get ΔG°rxn we have to use this formula when:
ΔG° = - RT ㏑ K
when ΔG° is Gibbs free energy
and R is the constant = 8.314 J/mol K
and T is the temperature in Kelvin = 25 °C+ 273 = 298 K
and when K = 4.4 x 10^-2
so, by substitution:
ΔG°= - 8.314 * 298 *㏑(4.4 x 10^-2)
= -7739 J = -7.7 KJ
b) then, to get E° cell for a redox reaction we have to use this formula:
ΔE° Cell = (RT / nF) ㏑K
when R is a constant = 8.314 J/molK
and T is the temperature in Kelvin = 25°C + 273 = 298 K
and n = no.of moles of e- from the balanced redox reaction= 3
and F is Faraday constant = 96485 C/mol
and K = 4.4 x 10^-2
so, by substitution:
∴ ΔE° cell = (8.314 * 298 / 3* 96485) *㏑(4.4 x 10^-2)
= - 2.7 x 10^-2 V
<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
I’m pretty sure it’s A) upper right!
Answer:
nah but ill take points tho
Explanation:
