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
Answer
Explanation:
Step1: Manipulate the given equations
The first equation is $P_4(s)+6Cl_2(g)\rightarrow4PCl_3(g)$ with $\Delta H_1 = - 2439\ kJ$. The second equation is $4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)$ with $\Delta H_2=3438\ kJ$. We want the reaction $PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$. Add the two given equations: $(P_4(s)+6Cl_2(g))+(4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g))$ gives $4PCl_5(g)\rightarrow4PCl_3(g) + 4Cl_2(g)$. The overall $\Delta H$ for this combined - equation is $\Delta H=\Delta H_1+\Delta H_2$.
Step2: Calculate $\Delta H$ for combined equation
$\Delta H=-2439\ kJ + 3438\ kJ=999\ kJ$. This is for the reaction $4PCl_5(g)\rightarrow4PCl_3(g)+4Cl_2(g)$.
Step3: Scale the reaction and $\Delta H$
Divide the reaction $4PCl_5(g)\rightarrow4PCl_3(g)+4Cl_2(g)$ by 4 to get $PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$. Divide the $\Delta H$ value by 4 as well. So $\Delta H=\frac{999\ kJ}{4}=249.75\approx250\ kJ$.
Answer:
- kJ