H2(g) + I2(g) ⇌ 2HI(g)
\(Q = \frac {[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}\)
\([\mathrm{HI}] = \frac {n_\mathrm{HI}}{V}\)
\(n_\mathrm{HI} = \frac {m_\mathrm{HI}}{M_\mathrm{HI}} = \frac {512\mathrm{g}}{(1,008 + 126,9)\mathrm{g/mol}} = 4,00287707\mathrm{mol}\)
\([\mathrm{HI}] = \frac {4,00287707\mathrm{mol}}{5,0\mathrm{dm^3}} = 0,80057541\mathrm{mol/dm^3}\)
\([\mathrm{H_2}] = \frac {n_\mathrm{H_2}}{V}\)
\(n_\mathrm{H_2} = \frac {m_\mathrm{H_2}}{M_\mathrm{H_2}} = \frac {1,01\mathrm{g}}{1,008 \cdot 2 \mathrm{g/mol}} = 0,50099206\mathrm{mol}\)
\([\mathrm{H_2}] = \frac {0,50099206\mathrm{mol}}{5,0\mathrm{dm^3}} = 0,10019841\mathrm{mol/dm^3}\)
\([\mathrm{I_2}] = \frac {n_\mathrm{I_2}}{V}\)
\(n_\mathrm{I_2} = \frac {m_\mathrm{I_2}}{M_\mathrm{I_2}} = \frac {127\mathrm{g}}{126,9 \cdot 2 \mathrm{g/mol}} = 0,50039401\mathrm{mol}\)
\([\mathrm{I_2}] = \frac {0,50039401\mathrm{mol}}{5,0\mathrm{dm^3}} = 0,10007880\mathrm{mol/dm^3}\)
\(Q = \frac {(0,80057541\mathrm{M})^2}{0,10019841\mathrm{M} \cdot 0,10007880\mathrm{M}} = 63,9148180 > K = 50\)
Eftersom \(Q > K\) kommer reaktionen att gå åt vänster.
