Kemi opgave 9.1-9.3 entalpi

Se evt: Entalpi.

Indholdsfortegnelse

Beregning på forbrænding af methan.

$LaTex: CH_4+2O_{2(g)}\rightarrow CO_{2(g)}+2H_2O_{(g)}$

Reaktanter:

Produkter:

$LaTex: \sum H^{\circ} \text{produkter} = -393,5\cdot 1 + -241,8\cdot 2 = -877,1$
$LaTex: \sum H^{\circ} \text{reaktanter} = -74,9$

$LaTex: \Delta H^{\circ} = \sum H^{\circ} \text{produkter}-\sum H^{\circ} \text{reaktanter} = -802,2 \frac{kJ}{mol}$

Derved kan vi konkludere at reaktionen er exoterm.

$LaTex: \frac{1000\frac{g}{kg}}{16,0425\frac{g}{mol}}\cdot -802,2\frac{kJ}{mol} =-50004,66 \frac{kJ}{kg}$

Delopgave 1: $LaTex: C_6 H_{12} O_6 \left( s \right) + 6O_2 \to 6CO_2 \left( g \right) + 6H_2 O\left( l \right) \\ \Delta H^{\circ} = 6 \cdot H^\ominus \left( {CO_2 \left( g \right)} \right) + 6 \cdot H^\ominus \left( {H_2 O\left( l \right)} \right) - H^\ominus \left( {C_6 H_{12} O_6 \left( s \right)} \right) + 6 \cdot H\left( {O_2 } \right) \\ 6 \cdot - 393.5 + 6 \cdot - 285.8 - \left( { - 1275 + 6 \cdot 0} \right) = - 2800.8\frac{{kJ}}{{mol}} \\ \frac{{1g}}{{180.156\frac{g}{{mol}}}} = 0.006mol \\ - 2800.8\frac{{kJ}}{{mol}} \cdot 0.006mol = - 16.805kJ \\$