consider the intermediate equations: c(s)+o2(g)→co2(g) δh1 = -393.5 kj 2co(g)+o2(g)→2co2(g) δh2 = -566.0 kj…

consider the intermediate equations: c(s)+o2(g)→co2(g) δh1 = -393.5 kj 2co(g)+o2(g)→2co2(g) δh2 = -566.0 kj 2h2o(g)→2h2(g)+o2(g) δh3 = 483.6 kj with the overall reaction: c(s)+h2o(g)→co(g)+h2(g) δhrxn =? what must be done to calculate the enthalpy of reaction? check all that apply. the first equation must be halved. the first equation must be reversed. the second equation must be halved. the second equation must be reversed. the third equation must be halved. the third equation must be reversed. what is the overall enthalpy of reaction? δhrxn = kj done

consider the intermediate equations: c(s)+o2(g)→co2(g) δh1 = -393.5 kj 2co(g)+o2(g)→2co2(g) δh2 = -566.0 kj 2h2o(g)→2h2(g)+o2(g) δh3 = 483.6 kj with the overall reaction: c(s)+h2o(g)→co(g)+h2(g) δhrxn =? what must be done to calculate the enthalpy of reaction? check all that apply. the first equation must be halved. the first equation must be reversed. the second equation must be halved. the second equation must be reversed. the third equation must be halved. the third equation must be reversed. what is the overall enthalpy of reaction? δhrxn = kj done

Answer

Explanation:

Step1: Analyze the reactants and products

We want to get $C(s)+H_2O(g)\rightarrow CO(g) + H_2(g)$ from the given equations. The first - equation $C(s)+O_2(g)\rightarrow CO_2(g)$ has $C(s)$ as a reactant which is correct as in the overall reaction. The second equation $2CO(g)+O_2(g)\rightarrow 2CO_2(g)$ has $CO$ as a product but we need it as a reactant in the overall reaction. The third equation $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ has $H_2O$ as a reactant and $H_2$ as a product which is correct for the overall reaction.

Step2: Manipulate the equations

  1. The first equation $C(s)+O_2(g)\rightarrow CO_2(g)$ can be used as is.
  2. For the second equation $2CO(g)+O_2(g)\rightarrow 2CO_2(g)$, we need to reverse it to get $CO$ on the product - side. Also, we need to halve it to get the correct stoichiometry of $CO$. So the new $\Delta H$ for the second - equation will be $\frac{566.0}{2}\ kJ$.
  3. For the third equation $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$, we need to halve it to get the correct stoichiometry of $H_2O$ and $H_2$. So the new $\Delta H$ for the third - equation will be $\frac{483.6}{2}\ kJ$.

Step3: Calculate the overall enthalpy

The overall enthalpy $\Delta H_{rxn}$ is calculated using Hess's law. [ \begin{align*} \Delta H_{rxn}&=\Delta H_1+\frac{\Delta H_2}{2}+\frac{\Delta H_3}{2}\ &=- 393.5+\frac{566.0}{2}+\frac{483.6}{2}\ &=-393.5 + 283+241.8\ &=131.3\ kJ \end{align*} ]

Answer:

The first equation must be used as is. The second equation must be halved and reversed. The third equation must be halved. $\Delta H_{rxn}=131.3\ kJ$