consider the following intermediate chemical equations.\n\n$p_4(s)+6cl_2(g)\rightarrow4pcl_3(g)$ $delta h_1…

consider the following intermediate chemical equations.\n\n$p_4(s)+6cl_2(g)\rightarrow4pcl_3(g)$ $delta h_1 = - 2,439kj$\n$4pcl_5(g)\rightarrow p_4(s)+10cl_2(g)$ $delta h_2 = 3,438kj$\n\nwhat is the enthalpy of the overall chemical reaction $pcl_5(g)\rightarrow pcl_3(g)+cl_2(g)$?\n\n-999 kj\n-250. kj\n250. kj\n999 kj

consider the following intermediate chemical equations.\n\n$p_4(s)+6cl_2(g)\rightarrow4pcl_3(g)$ $delta h_1 = - 2,439kj$\n$4pcl_5(g)\rightarrow p_4(s)+10cl_2(g)$ $delta h_2 = 3,438kj$\n\nwhat is the enthalpy of the overall chemical reaction $pcl_5(g)\rightarrow pcl_3(g)+cl_2(g)$?\n\n-999 kj\n-250. kj\n250. kj\n999 kj

Answer

Answer:

250 kJ

Explanation:

Step1: Manipulate equations

Divide the second - given equation by 4: $PCl_5(g)\rightarrow\frac{1}{4}P_4(s) + \frac{5}{2}Cl_2(g)$ with $\Delta H_{2}^{'}=\frac{3438}{4} = 859.5$ kJ. Divide the first - given equation by 4: $\frac{1}{4}P_4(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)$ with $\Delta H_{1}^{'}=\frac{- 2439}{4}=-609.75$ kJ.

Step2: Add the manipulated equations

Add the two new equations: $PCl_5(g)\rightarrow\frac{1}{4}P_4(s)+\frac{5}{2}Cl_2(g)$ and $\frac{1}{4}P_4(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)$. The $\frac{1}{4}P_4(s)$ terms cancel out, and we get $PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$.

Step3: Calculate the overall enthalpy

Use Hess's law. $\Delta H=\Delta H_{1}^{'}+\Delta H_{2}^{'}$. $\Delta H=-609.75 + 859.5=249.75\approx250$ kJ.