calculate $delta h^{circ}$ for the reaction $4hcl(g)+o_2(g)\rightarrow2cl_2(g)+2h_2o(g)$, using the…

calculate $delta h^{circ}$ for the reaction $4hcl(g)+o_2(g)\rightarrow2cl_2(g)+2h_2o(g)$, using the information provided below.\n$h_2(g)+cl_2(g)\rightarrow2hcl(g) delta h^{circ}=-185 kj$\n$2h_2(g)+o_2(g)\rightarrow2h_2o(g) delta h^{circ}=-483.7 kj$\n114 kj\n299 kj\n-299 kj\n-114 kj

calculate $delta h^{circ}$ for the reaction $4hcl(g)+o_2(g)\rightarrow2cl_2(g)+2h_2o(g)$, using the information provided below.\n$h_2(g)+cl_2(g)\rightarrow2hcl(g) delta h^{circ}=-185 kj$\n$2h_2(g)+o_2(g)\rightarrow2h_2o(g) delta h^{circ}=-483.7 kj$\n114 kj\n299 kj\n-299 kj\n-114 kj

Answer

Explanation:

Step1: Reverse the first - given reaction

Reverse $H_2(g)+Cl_2(g)\rightarrow2HCl(g)\ \Delta H^{\circ}=- 185\ kJ$ to get $2HCl(g)\rightarrow H_2(g)+Cl_2(g)\ \Delta H_1^{\circ}=185\ kJ$. Multiply this reversed reaction by 2 to match the amount of $HCl$ in the target reaction: $4HCl(g)\rightarrow2H_2(g)+2Cl_2(g)\ \Delta H_1^{\circ}=2\times185\ kJ = 370\ kJ$.

Step2: Keep the second - given reaction

Keep $2H_2(g)+O_2(g)\rightarrow2H_2O(g)\ \Delta H_2^{\circ}=-483.7\ kJ$.

Step3: Add the two modified reactions

Add the two reactions: [ \begin{align*} 4HCl(g)&\rightarrow2H_2(g)+2Cl_2(g)\ \Delta H_1^{\circ}=370\ kJ\ 2H_2(g)+O_2(g)&\rightarrow2H_2O(g)\ \Delta H_2^{\circ}=-483.7\ kJ\ \hline 4HCl(g)+O_2(g)&\rightarrow2Cl_2(g)+2H_2O(g)\ \Delta H^{\circ}=\Delta H_1^{\circ}+\Delta H_2^{\circ} \end{align*} ] Calculate $\Delta H^{\circ}=370\ kJ+( - 483.7\ kJ)=-114\ kJ$.

Answer:

-114 kJ