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3 years ago

What is the average of the two weights of a nickel and dime. Dime is 2.26g and nickel is 4.97g?

Chemistry

1 answer:

ANTONII [103]3 years ago

5 0

Dime=2.26g
Nickel=4.97g

$\\ \rm\Rrightarrow Avg\:Weight=\dfrac{Sum\:of\:Weights}{No\:of\:weights}$

$\\ \rm\Rrightarrow Avg\:weight=\dfrac{2.26+4.96}{2}$

$\\ \rm\Rrightarrow Avg\;weight=\dfrac{7.23}{2}$

$\\ \rm\Rrightarrow Avg\:Weight=3.6g$

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<u>Answer:</u> The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

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For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

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

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

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

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

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Putting values in above equation, we get:

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

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

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

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

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Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

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$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

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Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

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