Question

When a 11.90 L vessel containing 43.486 g of I2 is heated to 1162 K, some I2 dissociates: I2(g)→2I(g). If the final pressure in the vessel is 1.790 atm, A) what are the mole fractions of the two components I2(g) and I(g) after the reaction?

Answer 1

initial mol of I2 = mass of I2 / molar mass of I2
= 43.486 g/ 253.81 g/mol
= 0.171 mol

Let calculate the final number of moles using,
P*V = n*R*T
1.790 * 11.90 = n*0.0821*1162
n = 0.223 mol

I2 <----> 2I
0.171 0 (initial)
0.171-x 2x (final)

total moles = 0.171 - x + 2x = 0.171 + x mol

so,
0.171 + x = 0.223
x = 0.052

so, mol of I = 2*0.052 = 0.104 mol
moles of I2 = 0.171 -x = 0.171 - 0.052 = 0.119 mol

fraction of I = moles of I / total moles
= 0.104 / (0.104 +0.119)
= 0.47

fraction of I2 = 1- 0.47 = 0.53