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Gnoma [55]
2 years ago
13

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

Chemistry
1 answer:
Alekssandra [29.7K]2 years ago
4 0

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

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The quantum numbers for the last electron placed in three elements are listed below. Which of these is(are) NOT correct? Er (4 3
UNO [17]

Answer:

The three elements Erbium, Thallium and Osmium have incorrect quantum numbers for the last electron placed.

Explanation:

The 4 quantum numbers are (<em>n,l,ml,ms</em>):

  • <em>n</em> (Principal quantum number): it is the <u>number of the shell (level)</u> where the electron is placed.
  • <em>l </em>(Angular momentum quantum number or Secondary): it represents the <u>sublevel where the electron is</u> placed. There are 4 subleves: s, p d and f so secondary quantum number can take the number 0 (s), 1 (p), 2 (d) or 3 (f) depending on which sublevel the electron is placed.
  • <em>ml</em> (Magnetic quantum number):  it represents the <u>spatial orientation</u> of the electron <u>in respect of the sublevel the electron</u> is placed. For example: if the electron occupies the <em>s sublevel</em> the magnetic number will be <em>0</em>, if the electron occupies the <em>p sublevel</em> the magnetic number could be <em>-1,0,1</em>, if the electron occupies the <em>d sublevel</em> the magnetic number could be <em>-2,-1,0,1,2</em> and if the electron occupies the <em>f sublevel</em> the magnetic number could be <em>-3,-2,-1,0,1,2,3</em>. You can see this in the attachment related to the correct sublevel for the example.
  • <em>ms</em> (Spin quantum number): this number represents the possible rotation of the electron so it could be 1/2 (which is represented by an up arrow) or -1/2 (represented by an down arrow).

Let's analyze the last electron of each element. You can see the attachment for better understanding. The last electron it is represented with orange color.

- Erbium:

This element has 68 electrons so following the Moeller's Diagram to fill the the electronic configuration, we found that the last electron of Erbium it is in the <u>4th level </u>(shell), in the <u>f sublevel</u>. As Erbium has 12 electrons in the f sublevel, it is necessary to follow the Hund's rule (electrons must be placed singly in every sublevel before place a parallel electron) to placed correctly all of them. Finally, the last electron of Erbium stays in the middle of the sublevel and it is represented by a down arrow so the correct quantum numbers in the Erbium element are (4,3,1,-1/2).

- Thallium:

This element has 81 electrons and following the Moeller's Diagram, we found that it last electron it is in the <u>6th level</u>, in the <u>p sublevel</u>. As Thallium has 1 electron in the p sublevel, it is placed singly in the sublevel. So the last electron of Thallium it is represented by an up arrow so the correct quantum numbers in the Thallium element are (6,1,-1,1/2).

- Osmium:

Osmium has 76 electrons and following the steps  that we did with we the other elements, we noticed that its last electron it is in the <u>5th level</u>, in the <u>d sublevel</u>. Following the Hund's rule the last electron of Osmium has a magnetic quantum number of -2 and its spin quantum number is -1/2, so the quantum numbers in the Osmium element are (5,2,-2,-1/2).

<u>Note:</u>

- Remember that the <em>s sublevel</em> has place for 2 electrons, the <u>p sublevel</u> has place for 6 electrons, the <u>d sublevel</u> has place for 10 electrons and the<em> f sublevel</em> has place for 14 electrons.

3 0
3 years ago
In a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes. Determine the average relative atomic
Burka [1]

The average relative atomic mass of a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes is 9.6g.

<h3>What is Atomic mass? </h3>

Atomic mass is defined as the whole mass of an atom.

It is also defined as the sum of atomic number and number of neutrons.

Atomic mass = Atomic number + neutrons

<h3>What is Isotopes?</h3>

Isotopes are the those element which have same atomic number but have different mass number and number of neutrons.

The average relative atomic mass can be calculated as

mass of isotopes/ mass of sample

mass of all isotopes = (70 × 9X) + (30 × 11X)

=(630 + 330) X

= 960X

Average relative atomic mass = 960X/ 100 X

= 9.6 g

Thus, we concluded that the average relative atomic mass of a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes is 9.6g.

learn more about atomic mass:

brainly.com/question/14250653

#SPJ9

5 0
1 year ago
The decay of a living things allows chemical elements to be lost.
Lelu [443]

Answer: is is true

Explanation:

8 0
1 year ago
How many grams of NaCl are needed to prepare 50.0 grams of 35.0% of salt solution?
nevsk [136]

Answer:

17.5 g

Explanation:

Given data

  • Mass of solution to be prepared: 50.0 grams
  • Concentration of the salt solution: 35.0%

The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.

50.0gSolution \times \frac{35.0gNaCl}{100gSolution} = 17.5gNaCl

4 0
3 years ago
A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution?
melomori [17]
[OH-]: <span>0.0063095734448</span>
7 0
3 years ago
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