What is the variable for the equilibrium constant Re K Ec E

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

2 years ago

I need help with all of it pls and quick

Mazyrski [523]

With what are you need help?

5 0

2 years ago

Question 10 of 25

kap26 [50]

Answer:

Correct option would be A. Mg(s) → Mg2+ + 2e-

Explanation:

Marked as correct answer on Quiz ;) (A P E X)

7 0

2 years ago

Read 2 more answers

At 19 °C, the vapor pressure of pure ethanol is 40.0 mmHg. What is the pressure in atm? atm

m_a_m_a [10]

Answer:

0.05263158 atm

Explanation: mmHg to Atmosphere Conversion Example. Task: Convert 975 mmHg to atmospheres (show work) Formula: mmHg ÷ 760 = atm Calculations: 975 mmHg ÷ 760 = 1.28289474 atm Result: 975 mmHg is equal to 1.28289474 atm. This is an example for how i got the answer.

5 0

1 year ago

5. A standardized mass was known to weigh exactly 5.0000g. It was weighed on a newly

sergey [27]

Answer:

Absolute error $= 0.001$