a & b.
N2(g) + O2(g) ⇌ 2NO(g)
\[\begin{aligned}Q &= \frac {[\mathrm{NO}]^2}{[\mathrm{N_2}][\mathrm{O_2}]} = \frac {(0,0300\mathrm{M})^2}{0,810\mathrm{M} \cdot 0,750\mathrm{M}} = \\ &= 0,00148148 < K = 0,0025\end{aligned}\]
Eftersom \(Q < K\) är systemet inte i jämvikt, utan kommer att gå åt höger.
c.
|
| [N2] | [O2] | [NO] |
|
| f.r. | \[0,810\] | \[0,750\] | \[0,0300\] | M |
| ∆ | \[-x\] | \[-x\] | \[+2x\] | M |
| v.j. | \[0,810-x\] | \[0,750-x\] | \[0,0300+2x\] | M |
\[\begin{aligned}K &= \frac {[\mathrm{NO}]^2}{[\mathrm{N_2}][\mathrm{O_2}]} \\ 0,0025 &= \frac {(0,0300 + 2x)^2}{(0,810-x)(0,750-x)}\end{aligned}\]
\[\begin{aligned}0,0025 &= \frac {0,0009 + 0,12x + 4x^2}{0,6075 -1,56x + x^2} \\ 0,0025(0,6075 -1,56x + x^2) &= 0,0009 + 0,12x + 4x^2 \\ 0,00151875 - 0,0039x + 0,0025x^2 &= 0,0009 + 0,12x + 4x^2 \\ 0 &= -0,00061875 + 0,1239x + 3,9975x^2 \\ x^2 + \frac {0,1239}{3,9975}x - \frac {0,00061875}{3,9975} &= 0\end{aligned}\]
\(pq\)-formeln ger:
\[\begin{aligned}x &= 0,00437609 \\ [\mathrm{N_2}] &= (0,810 - x)\mathrm{M} = (0,810 - 0,00437609)\mathrm{M} = \\ &= 0,80562391\mathrm{M} ≈ 0,806\mathrm{M} \\ [\mathrm{O_2}] &= (0,750 - x)\mathrm{M} = (0,750 - 0,00437609)\mathrm{M} = \\ &= 0,74562391\mathrm{M} ≈ 0,746\mathrm{M} \\ [\mathrm{NO}] &= (0,0300 + 2x)\mathrm{M} = (0,0300 - 2 \cdot 0,00437609)\mathrm{M} = \\ &= 0,03875218\mathrm{M} ≈ 0,0388\mathrm{M}\end{aligned}\]
