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Ratling [72]
3 years ago
11

Calculate Δ H° for the reaction C 4H 4( g) + 2H 2( g) → C 4H 8( g), using the following data: Δ H° combustion for C 4H 4( g) = –

2341 kJ/mol Δ H° combustion for H 2( g) = –286 kJ/mol Δ H° combustion for C 4H 8( g) = –2755 kJ/mol
Chemistry
1 answer:
arlik [135]3 years ago
5 0

<u>Answer:</u> The enthalpy of the reaction is coming out to be 2231 kJ.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_{(product)}]-\sum [n\times \Delta H^o_{(reactant)}]

For the given chemical reaction:

C_4H_4(g)+2H_2(g)\rightarrow C_4H_8(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_{(C_4H_8(g))})]-[(1\times \Delta H^o_{(C_4H_4(g))})+(2\times \Delta H^o_{(H_2(g))})]

We are given:

\Delta H^o_{(C_4H_8(g))}=-2755kJ/mol\\\Delta H^o_{(H_2(g))}=-286kJ/mol\\\Delta H^o_{(C_4H_4(g))}=-2341kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-2755))]-[(1\times (-286))+(2\times (-2341))]\\\\\Delta H^o_{rxn}=2213kJ

Hence, the enthalpy of the reaction is coming out to be 2231 kJ.

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                = 3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}  

                 = 3.03975 \times 10^{5} Pa

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                      = 3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})

                      = 1823.85 Nm

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As internal energy of the gas \Delta E is as follows.

                     \Delta E = Q - W

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                                  = -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.

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