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3 years ago

Determine the enthalpy change of the following reaction: CO + H2O -> H2 + CO2

Chemistry

2 answers:

Tomtit [17]3 years ago

8 0

I think A is correct:

deltaH=(Enthalpie of reactents)-(Enthalpie of products)=110.525+285.5-393.5=2.525~2.825

Paraphin [41]3 years ago

7 0

Answer:

A. The enthalpy change for the reaction is 2.825 kJ

Explanation:

The given reaction is:

CO + H2O → H2 + CO2

The enthalpy change for a reaction is given as:

$\Delta H = \sum n(p)\Delta H_{f}^{0}(products)-\sum n(r)\Delta H_{f}^{0}(reactants)$

where np and nr are the number of moles of products and reactants

ΔH⁰f are the standard enthalpies of formation of the respective reactants and products

$\Delta H = [1\Delta H_{f}^{0}(H2)+1\Delta H_{f}^{0}(CO2)]-[1\Delta H_{f}^{0}(CO)+1\Delta H_{f}^{0}(H2O)]$

Substituting the given enthalpy data:

ΔH = [1(0) + 1(-393.5)] - [1(-110.525) + 1(-285.8)] = 2.825 kJ

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andrew11 [14]

The given question is incomplete. The complete question is:

Consider the spontaneous dissolution of NaCl in water: NaCl(s) → Na+(aq) + Cl −(aq) Predict whether the entropy change for this process is positive or negative and explain your answer. Then, predict whether the free energy change for the process is positive or negative and explain your answer.

Answer: The entropy change is positive. The free energy change for the process is negative and reaction is spontaneous.

Explanation:

Entropy is defined as the degree of randomness of a system. For a system in which randomness increses , the enetropy is said to increase and $\Delta S$ is positive.

As in the given reaction : $NaCl(s)\rightarrow Na^+(aq)+Cl^-(aq)$ , the solid is dissociating to give ions , randomness increases and thus entropy is positive.

Also the dissociation of a molecule requires energy , thus the enthalpy $\Delta H$ is also positive as heat is absorbed by the system.

$\Delta G=\Delta H-T\Delta S$

$\Delta G=(+ve)-T(+ve)$

$\Delta G=-ve$ when Temperature is high.

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Rzqust [24]

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