Calculate the equilibrium constant for the following two reactions. the equilibrium concentration of each reactant and product is given in the box below the reaction. rank the extent of each reaction at equilibrium by calculating the equilibrium constant. note that the "extent of each reaction" refers to how much reactant is converted to product at equilibrium.
reaction 1: reaction 2:
co(g)+o2(g)⇌2co2(g) 2n2(g)+o2(g)⇌2no2(g)
equilibrium concentrations equilibrium concentrations
[co]=1.3m [n2]=1.5m
[o2]=1.3m [o2]=2.1m
[co2]=8.8m [no2]=0.04m
compare kc values to indicate which reaction proceeds to a greater extent at equilibrium.
view available hint(s)
compare values to indicate which reaction proceeds to a greater extent at equilibrium.
reaction 2, with kc=45.8> reaction 1, with kc=3.4×10−4
reaction 1, with kc=5.21> reaction 2, with kc=1.3×10−2
reaction 2, with kc=3000> reaction 1, with kc=2.2×10−2
reaction 1, with kc=45.8> reaction 2, with kc=3.4×10−4

Reaction 1: Equilibrium constant (Kc) = 45.82

Reaction 2: Equilibrium constant (Kc) = 3.4×10⁻⁴

Thus Hint number 4 is correct: Reaction 1, with Kc=45.8 proceed to a greater extent than Reaction 2 with Kc=3.4×10⁻⁴

Explanation:

Reaction 1: CO(g)+O2(g)⇄2CO2(g)

Kc = [CO2]²/[CO][O2] = (8.8)²/(1.3 x 1.3) = 45.82, this shows that there are more products that reactant at equilibrium

Reaction 2: 2N2(g)+O2(g)⇄2NO2(g)

Kc = [NO2]²/[N2]²[O2] = (0.04)²/[(1.5)²(2.1)] = 3.4×10⁻⁴ which is less than 1, showing that the denominator in the equation is higher than the numerator thus more reactants are present than products at equilibrium.

the 2 Kc show that reaction 1 proceeds to a greater extent than reaction 2 at equilibrium.

hints number 4.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm