Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equations: C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1,123 kJ C(s) + O2(g) ---> CO2(g) ΔH° = -340 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -211 kJ

Answers

Answer 1
Answer:

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

C_2H_2(g) + (5)/(2)O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_(1) = -1,123 kJ...[1]

C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_(2) = -340 kJ..[2]

H_2(g) + (1)/(2)O_2(g)\rightarrow H_2O(l) ,\Delta H^o_(3) = -211 kJ..[3]

2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_(4) =?..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

\Delta H^o_(4)=2* \Delta H^o_(2)+\Delta H^o_(3) - \Delta H^o_(1)

\Delta H^o_(4)=2* (-340 kJ) + (-211 kJ) - (-1,123 kJ)

\Delta H^o_(4)=232 kJ

The enthalpy for given reaction is 232 kilo Joules.


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Consider the reaction: 2 HI (g) ⇌ H₂ (g) + I₂ (g) At equilibrium, the partial pressure of HI is 1.9 atm and the partial pressures of H₂ and I₂ are 7.9 and 2.3 respectively. What is Kp for this equilibrium?

Answers

Answer:

To determine the value of Kp for the given equilibrium, we need to use the partial pressures of the gases involved.

In the balanced equation: 2 HI (g) ⇌ H₂ (g) + I₂ (g), the stoichiometric coefficients are 2, 1, and 1 respectively.

At equilibrium, the expression for Kp is given by:

Kp = (P(H₂) * P(I₂)) / (P(HI)²)

Using the provided partial pressures:

P(HI) = 1.9 atm

P(H₂) = 7.9 atm

P(I₂) = 2.3 atm

Substituting these values into the expression for Kp:

Kp = (7.9 * 2.3) / (1.9²)

Kp ≈ 19.5 / 3.61

Calculating the result:

Kp ≈ 5.4

Therefore, the value of Kp for the given equilibrium is approximately 5.4.

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Compare and contrast the characteristics of metals and nonmetals.

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metals are hard and no metals are liquid

What is an extensive property that can be calculated?

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Answer: The property which depends on the quantity of the substance is called an extensive property. The free energy change for a reaction (Δ G) depends on the quantity of the substance and is therefore an extensive property. It shows the additive nature. The extensive property Δ G is easily calculated from the formula, ΔG = -nFE cell.

Explanation:

Final answer:

An extensive property is one that changes when the size of the sample changes. One such property that can be calculated is enthalpy. Enthalpy can be calculated using the formula H = E + PV.

Explanation:

An extensive property is a property that changes when the size of the sample changes. Examples include mass, volume, length, and total charge. One extensive property that can be calculated is enthalpy.

The enthalpy of a system can be calculated using the formula H = E + PV, where H represents the enthalpy, E the internal energy of the system, P the pressure, and V the volume. Like other extensive properties, the enthalpy of a system would change with the quantity or size of the sample.

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Round the following number to three significant figures: 156.5021 L

Answers

you round the 6 to a seven because there is a five following the decimal157

Please answer my question. #6-10: Consider the energy diagram below:
6. What is the activation energy for this reaction? _____
7. What is the change in free energy for the reaction? _____
8. How many intermediates are involved in this reaction? _____
9. How many transition states are involved in the reaction mechanism? _____
10. Is the reaction endergonic or exergonic overall?

Answers

6. Activation energy is the difference in energy between the reactant and the transition state. The energy of the reactant and the transition state are 5 kJ/mol and 15 kJ/mol respectively. The activation energy for this reaction is 10 kJ/mol.

7. The change in free energy for the reaction is the difference in energy between the reactant and the product. The energy of the reactant and the product are 5 kJ/mol and 10 kJ/mol respectively. The change in free energy for the reaction is 5 kJ/mol.

8. There are no intermediates involved in this reaction.

9.  There is only one transition state involved in the reaction mechanism. The transition state is indicated by the highest point of the graph.

10. The reaction is endergonic overall. The energy of the product is higher than the energy of the reactants, which is only possible if energy is absorbed by the reaction.

Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.61×10−11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. What is the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.180 M NaOH solution? Express your answer numerically as the ratio of molar solubility in H2O to the molar solubility in NaOH.

Answers

Answer:

The ration of the molar solubility is 165068.49.

Explanation:

The solubility reaction of the magnesium hydroxide in the pure water is as follows.

Mg(OH)_(2)\Leftrightarrow Mg^(2+)(aq)+2(OH)^(-)(aq)

              [Mg^(2+)][OH^(-)]

Initial      0          0

Equili     +S       +2S

Final      S          2S

K_(sp)=[Mg^(2+)][OH^(-)]

5.61* 10^(-11)=(S)(2S)^(2)

S=((5.61* 10^(-11))/(4))^(1/3)=2.41* 10^(-4)M

Solubility of Mg(OH)_(2) in 0.180 M NaOH is a follows.

Mg(OH)_(2)\Leftrightarrow Mg^(2+)(aq)+2(OH)^(-)(aq)

              [Mg^(2+)][OH^(-)]

Initial      0          0

Equili     +S       +2S

Final      S          2S+0.180M

K_(sp)=[Mg^(2+)][OH^(-)]

5.61* 10^(-11)=(S)(2S+0.180)^(2)

S=1.46* 10^(-9)M

Ratio\,of\,solubility=(2.41* 10^(-4))/(1.46* 10^(-9))=165068.49

Therefore, The ration of the molar solubility is 165068.49.