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3 years ago

I need help with these two

Chemistry

1 answer:

bearhunter [10]3 years ago

3 0

Answer:

6. d, 7. a

Explanation:

6. Molarity is a number of moles solute in 1 L solution.

7. 1 L solution - 2.5 mol K2CO3

20 L - x mol K2CO3

x =20*2.5/1 = 50 mol K2CO3

Molar mass(KCO3) = M(K) + M(C) + 3M(O)= 39 +12 +3*16= 99 g/mol

99 g/mol *50 mol = 4950 g KCO3 Closest answer is A.

Actually KCO3 does not exist, in reality it should be K2CO3.

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

We are given:

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

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

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

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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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