consider the following intermediate chemical equations.\n\n$p_4(s)+3o_2(g)\\rightarrow p_4o_6(s)$ $delta…

consider the following intermediate chemical equations.\n\n$p_4(s)+3o_2(g)\\rightarrow p_4o_6(s)$ $delta h_1=-1,640kj$\n$p_4o_{10}(s)\\rightarrow p_4(s)+5o_2(g)$ $delta h_2 = 2,940.1kj$\n\nwhat is the enthalpy of the overall chemical reaction $p_4o_6(s)+2o_2(g)\\rightarrow p_4o_{10}(s)$?\n\n-4,580 kj\n-1,300 kj\n1,300 kj\n4,580 kj

consider the following intermediate chemical equations.\n\n$p_4(s)+3o_2(g)\\rightarrow p_4o_6(s)$ $delta h_1=-1,640kj$\n$p_4o_{10}(s)\\rightarrow p_4(s)+5o_2(g)$ $delta h_2 = 2,940.1kj$\n\nwhat is the enthalpy of the overall chemical reaction $p_4o_6(s)+2o_2(g)\\rightarrow p_4o_{10}(s)$?\n\n-4,580 kj\n-1,300 kj\n1,300 kj\n4,580 kj

Answer

Explanation:

Step1: Reverse the first equation

When we reverse the equation $P_4(s)+3O_2(g)\rightarrow P_4O_6(s)$ with $\Delta H_1 = - 1640kJ$, the new equation is $P_4O_6(s)\rightarrow P_4(s)+3O_2(g)$ and the new $\Delta H$ value is $\Delta H_{1r}=1640kJ$.

Step2: Add the reversed - first and the second equation

The second equation is $P_4O_{10}(s)\rightarrow P_4(s)+5O_2(g)$ with $\Delta H_2 = 2940.1kJ$. Adding $P_4O_6(s)\rightarrow P_4(s)+3O_2(g)$ ($\Delta H_{1r}=1640kJ$) and $P_4(s)+5O_2(g)\rightarrow P_4O_{10}(s)$ (reverse of the second equation with $\Delta H_{2r}=- 2940.1kJ$) gives $P_4O_6(s)+2O_2(g)\rightarrow P_4O_{10}(s)$. The overall $\Delta H=\Delta H_{1r}+\Delta H_{2r}=1640+( - 2940.1)=-1300.1\approx - 1300kJ$.

Answer:

-1,300 kJ