The combustion of white phosphorous $(P_4(s))$ liberates $1544\;kJ$ for each mole of formed $P_2O_5(s)$ , the combustion of red phosphorous $(P(s))$ $1509\;kJ$. Calculate the standard enthalpy of the following reaction: (c) $P_4(s)$ $\longrightarrow$ $4P(s)$ $\Delta H_c = ?$
(a) $\frac{1}{2}P_4(s)$ $+$ $\frac{5}{2}O_2(g)$ $\longrightarrow$ $\color{red}1\color{black}P_2O_5(s)$ $\Delta H_a$ $=$ $\color{red}-\color{black}1544kJ$ (b) $2P(s)$ $+$ $\frac{5}{2}O_2(g)$ $\longrightarrow$ $\color{red}1\color{black}P_2O_5(s)$ $\Delta H_b$ $=$ $\color{red}-\color{black}1509kJ$
2(a) $P_4(s)$ $+$ $5O_2(g)$ $\longrightarrow$ $2P_2O_5(s)$ $2\Delta H_a$ -2(b) $2P_2O_5(s)$ $\longrightarrow$ $4P(s)$ $+$ $5O_2(g)$ $-2\Delta H_b $
2(a)-2(b) $P_4(s)$ $+$ $5O_2(g)$ $+$ $2P_2O_5(s)$ $\longrightarrow$ $2P_2O_5(s)$ $+$ $4P(s)$ $+$ $5O_2(g)$ $2\Delta H_a$ $-$ $2\Delta H_b$ (c)= 2(a)-2(b) $P_4(s)$ $\longrightarrow$ $4P(s)$ $\Delta H_c$ $=$ $2\Delta H_a-2\Delta H_b$ $=$ $- 70\;kJ $
This reaction is exothermic!