consider the following intermediate chemical equations. c(s)+o2(g)→co2(g) δh1 = -393.5kj 2co(g)+o2(g)→2co2(g)…
consider the following intermediate chemical equations. c(s)+o2(g)→co2(g) δh1 = -393.5kj 2co(g)+o2(g)→2co2(g) δh2 = -566.0kj 2h2o(g)→2h2(g)+o2(g) δh3 = 483.6kj the overall chemical equation is c(s)+h2o(g)→co(g)+h2(g). to calculate the final enthalpy of the overall chemical equation, which step must occur? reverse the first equation, and change the sign of the enthalpy. then, add. reverse the second equation, and change the sign of the enthalpy. then, add. multiply the first equation by three, and triple the enthalpy. then, add. divide the third equation by two, and double the enthalpy. then, add.
Answer
Explanation:
Step1: Analyze the given equations
We have three intermediate - equations and an overall equation. We need to manipulate the intermediate equations to get the overall equation using Hess's law.
Step2: Analyze each option
- Option 1: Reversing the first equation $C(s)+O_2(g)\rightarrow CO_2(g)\ \Delta H_1 = - 393.5kJ$ gives $CO_2(g)\rightarrow C(s)+O_2(g)\ \Delta H=-(-393.5kJ) = 393.5kJ$. This is not helpful for getting the overall equation $C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)$.
- Option 2: Reverse the second equation $2CO(g)+O_2(g)\rightarrow 2CO_2(g)\ \Delta H_2=-566.0kJ$. The reversed equation is $2CO_2(g)\rightarrow 2CO(g)+O_2(g)\ \Delta H = 566.0kJ$. When we combine this with the other equations, we can manipulate them to get the overall equation.
- Option 3: Multiplying the first equation by three gives $3C(s)+3O_2(g)\rightarrow 3CO_2(g)\ \Delta H = 3\times(-393.5kJ)$. This will not lead to the overall equation.
- Option 4: Dividing the third equation $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)\ \Delta H_3 = 483.6kJ$ by two gives $H_2O(g)\rightarrow H_2(g)+\frac{1}{2}O_2(g)\ \Delta H=\frac{483.6kJ}{2}=241.8kJ$. This alone will not lead to the overall equation.