Question

consider the following intermediate chemical equations.

$$\ce{p4(s) + 3o2(g) -> p4o6(s)} \quad \delta h_1 = -1,640\\,kj$$
$$\ce{p4o10(s) -> p4(s) + 5o2(g)} \quad \delta h_2 = 2,940\\,kj$$

what is the enthalpy of the overall chemical reaction $\ce{p4o6(s) + 2o2(g) -> p4o10(s)}$?

• $-4,580\\,kj$
• $-1,300\\,kj$
• $1,300\\,kj$
• $4,580\\,kj

Explanation:

Step1: Reverse the first equation

Reverse \( \ce{P4(s) + 3O2(g) -> P4O6(s)} \) to get \( \ce{P4O6(s) -> P4(s) + 3O2(g)} \), and change \( \Delta H_1 \) to \( +1640\space kJ \) (since reversing a reaction changes the sign of \( \Delta H \)).

Step2: Reverse the second equation

Reverse \( \ce{P4O10(s) -> P4(s) + 5O2(g)} \) to get \( \ce{P4(s) + 5O2(g) -> P4O10(s)} \), and change \( \Delta H_2 \) to \( - 2940\space kJ \) (reversing the reaction flips the sign of \( \Delta H \)).

Step3: Add the two reversed equations

First reversed equation: \( \ce{P4O6(s) -> P4(s) + 3O2(g)} \), \( \Delta H = + 1640\space kJ \)
Second reversed equation: \( \ce{P4(s) + 5O2(g) -> P4O10(s)} \), \( \Delta H=-2940\space kJ \)
Adding them: \( \ce{P4O6(s) + 2O2(g) -> P4O10(s)} \)
Calculate the total \( \Delta H \): \( \Delta H = 1640 - 2940=-1300\space kJ \)

Answer:

-1,300 kJ (the option with -1,300 kJ)