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Elena-2011 [213]
2 years ago
14

Study the reactions.

Chemistry
1 answer:
yulyashka [42]2 years ago
3 0

Answer:

\displaystyle \Delta H = -2426\text{ kJ}

Explanation:

To find the enthalpy change of the target reaction, we can use Hess's Law.

Reversing the third reaction yields:

\displaystyle \text{C$_3$H$_8$(g)} \longrightarrow 3\text{C(s)} + 4\text{H$_2$(g)}\;\;\;\;\;\Delta H = -106\text{ kJ}

Multiplying the first reaction by three yields:

\displaystyle 3\text{C(s)} + 3\text{O$_2$(g)} \longrightarrow 3\text{CO$_2$}(g)}\;\;\;\;\; \Delta H = -1.18\times 10^3\text{ kJ}

Multiplying the second reaction by four yields:

\displaystyle 4\text{H$_2$(g)} + 2\text{O$_2$(g)} \longrightarrow 4\text{H$_2$O($\ell$)} \;\;\;\;\;  \Delta H = -1.14\times 10^3 \text{ kJ}

Adding all equations yield:

\displaystyle \text{C$_3$H$_8$(g)} + 5\text{O$_2$(g)} \longrightarrow 3\text{CO$_2$(g)} + 4\text{H$_2$O($\ell$)} \;\;\;\;\; \Delta H = -2426\text{ kJ}

Hence, the enthalpy change of the target reaction is -2426 kJ.

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<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

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

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\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

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\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

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