consider the chemical equations shown here.\np4(s) + 3o2(g) → p4o6(s) δh1 = -1,640.1 kj\np4o10(s) → p4(s) +…
consider the chemical equations shown here.\np4(s) + 3o2(g) → p4o6(s) δh1 = -1,640.1 kj\np4o10(s) → p4(s) + 5o2(g) δh2 = 2,940.1 kj\nwhat is the overall enthalpy of reaction for the equation shown below?\nround the answer to the nearest whole number.\np4o6(s) + 2o2(g) → p4o10(s)
Answer
Explanation:
Step1: Revertir la primera ecuación
$$ \begin{align*} P_4O_6(s)&\to P_4(s)+3O_2(g)\ \Delta H_1'& = 1640.1\ kJ \end{align*} $$
Step2: Revertir la segunda ecuación
$$ \begin{align*} P_4(s)+5O_2(g)&\to P_4O_{10}(s)\ \Delta H_2'&=- 2940.1\ kJ \end{align*} $$
Step3: Sumar las ecuaciones revertidas
$$ \begin{align*} P_4O_6(s)+P_4(s)+5O_2(g)&\to P_4(s)+3O_2(g)+P_4O_{10}(s)\ P_4O_6(s)+2O_2(g)&\to P_4O_{10}(s) \end{align*} $$ $$ \begin{align*} \Delta H&=\Delta H_1'+\Delta H_2'\ &=1640.1+( - 2940.1)\ &=-1300\ kJ \end{align*} $$
Answer:
$-1300$