Question

The initial amounts are given for the reactants, and a final observed concentration for the product. If these amounts are sufficient to permit equilibrium to be reached (you must determine this) calculate Keq for the reaction at this temperature. (Hint: notice the phase of each participant.) Assume V = 1.0 L

Balanced chemical equation: 6 H2 (g) + P4 (s) <-> 4 PH3 (g)

initial amounts: (6 H2) 3.86 x 10^(−3) / (P4)M 0.500 g/ (4PH3) 6.05 x 10^(−5)M

equilibrium concentration: 1.27 x 10^(−4) M

Answer 1

Given, Volume = 1.0 L

6 H2 (g) + P4 (s) <-----------------> 4 PH3 (g)

6 H2 (g) + P4 (s) <-----------------> 4 PH3 (g)

Initial concentration : 3.86 *10^-3 M 0.500 g 6.05* 10^-5 M

final concentration : 1.27 * 10^-4 M

PH3 produced through the reaction = final concentration - initial concentration

= 1.27 * 10^-4 M - 6.05* 10^-5 M

= 6.605* 10^-5 M

Therefore, additional no.of moles of PH3 = Concentration * Volume = 6.605 * 10^-5 M * 1.000L = 6.605 * 10^-5 moles

From the stoichiometric equation,

  4 moles of PH3 requires ---------------->   6 moles of H2 reacts to produces

==> 6.605 * 10^-5 M of PH3 requires ---------------> (6/4 ) * 6.605* 10^-5 moles of H2

= 9.908 * 10^-5 of moles of H2

( i.e., 9.908 * 10^-5 of moles of H2 are consumed during the reaction )

Molarity of 9.908 * 10^-5 of moles of H2 = 9.908 * 10^-5 / 1.0 L = 9.908 * 10^-5 M

Therefore,

final concentration of H2 = intial concentration - reacted concentration = 3.86 * 10^-3 - 9.908 * 10^-5

==> [H2] = 3.76 * 10^-3 M

Now, Equillibrium constant = K_eq = [ PH3 ]⁴ / [H2]⁶ where, [ x] = concentration of x at equillibrium

( since, P4 (s) is a solid ,we don't consider it's concentration)

= [1.27 * 10^-4 ] ⁴ / [3.76 * 10^-3 ] ⁶

= 0.0921 L² mol^-2

= 9.21 * 10^-2 L² mol^-2

Therefore, K_eq = 9.21 * 10^-2 L² mol^-2