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# Calculate values of K, ΔG° (in kJ), and ΔS° (in J/K). 2.56 Given:atm of A2(g) and 2.56 atm of B2(g) were mixed in a 1.00 L cylinder at 25°C.?

Consider the reaction

A2(g) + B2(g) equilibrium reaction arrow 2 C(g) where ΔH° = −104.3 kJ.

In a particular experiment, 2.56 atm of A2(g) and 2.56 atm of B2(g) were mixed in a 1.00 L cylinder at 25°C and allowed to reach equilibrium. Then the molecules of A2(g) were counted by using a very sensitive technique, and 2.82 ✕ 1013 molecules were found. For this reaction, calculate the values of K, ΔG° (in kJ), and ΔS° (in J/K).

I tried using the molecules found into finding pressure but I just do not understand how I am to use that to find the values I need.

### 1 Answer

- jacob sLv 71 month ago
P = 2.58atm , T = 25 C = 298 K , V = 1L

we find moles of A2 we use PV = nRT

2.58 atmx 1L = n x 0.0821 Latm/molK x 298 K

n = 0.1055 mol , [A2] = mol / vol = ( 0.1055 mol/1L) = 0.1055 M

at equilibirum A2 molecules = 2.84 x 10^ 13 , A2 mol = ( 2.84 x 10^ 13 / 6.022 x 10^ 23) = 4.716 x 10^ -11 mol

[A2] = mol / vol = ( 4.716 x 10^ -11 mol / 1L) = 4.716 x 10^ -11 M

at equilibirum [A2] = equilibirum [B2] = 4.716 x 10^ -11 M which is nearly 0. Hence nearly all A2 converted to C

[C] = 2 times [A2] initial = 2 x 0.1055 = 0.211 M

K = [C]^2 / [A2] [B2] = ( 0.211 ) / ( 4.716 x 10^ -11) ( 4.716 x 10^ -11)

= 9.5 x 10^ 19

dG = - RT ln K = - 8.314 J/K x 298K x ln ( 9.5 x 10^ 19)

= - 113969 J/mol = - 114 KJ

dS = (dH - dG ) / T = ( - 104.3 KJ - ( -114 KJ) / ( 298K) = 0.03255 KJ/mol = 32.55 KJ/mol