At 350°c, keq = 1.67 × 10-2 for the reversible reaction 2hi (g) ⇌ h2 (g) + i2 (g). what is the concentration of hi at equilibrium if [ h2 ] is 2.44 × 10-3 m and [ i2 ] is 7.18 × 10-5 m?

According to the reversible reaction equation:

2Hi(g) ↔ H2(g) + i2(g)

and when Keq is the concentration of the products / the concentration of the reactants.

Keq = [H2][i2]/[Hi]^2

when we have Keq = 1.67 x 10^-2

[H2] = 2.44 x 10^-3

[i2] = 7.18 x 10^-5

so, by substitution:

1.67 x 10^-2 = (2.44 x 10^-3)*(7.18x10^-5)/[Hi]^2

∴[Hi] = 0.0033 M

500bc indiais the answer.

Idon’t know what the answer is i wish i could