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3 years ago

1. What is the frequency of the following:

Chemistry

1 answer:

den301095 [7]3 years ago

5 0

Answer:

its soooooo ezy

Explanation:

82828I 87is 818373773622771829383746171948468328933737373737378364737272828374736367181i3i46738100104848070792761527484930201834728the 8585949476425675666

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Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant at for the following reaction. N2(g)H

soldier1979 [14.2K]

Answer: The equilibrium constant for this reaction is $5.85\times 10^{5}$

Explanation:

The equation used to calculate standard Gibbs free change is of a reaction is:

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

For the given chemical reaction:

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

The equation for the standard Gibbs free change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_{(NH_3(g))})]-[(1\times \Delta G^o_{(N_2)})+(3\times \Delta G^o_{(H_2)})]$

We are given:

$\Delta G^o_{(NH_3(g))}=-16.45kJ/mol\\\Delta G^o_{(N_2)}=0kJ/mol\\\Delta G^o_{(H_2)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol$

To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -32.9 kJ/mol = -35900 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = $25^oC=[273+25]K=298K$

$K_{eq}$ = equilibrium constant at 25°C = ?

Putting values in above equation, we get:

$-32900J/mol=-(8.314J/Kmol)\times 298K\times \ln K_{eq}\\\\K_{eq}=e^{13.279}=5.85\times 10^{5}$

Hence, the equilibrium constant for this reaction is $5.85\times 10^{5}$

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3 years ago