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crimeas [40]
3 years ago
14

A reaction has a standard free-energy change of -14.50 kJ mol(-3.466 kcal mol). Calculate the equilibrium constant for the react

ion at 25 °C. 5.85 K
Chemistry
1 answer:
den301095 [7]3 years ago
5 0

Answer:  The equilibrium constant for the reaction at 25 °C is 346.7

Explanation:

Formula used :

\Delta G^o=-2.303\times RT\times \log K_c

where,

\Delta G^o = standard Gibb's free energy change = -14.50kJ/mol =14500 J/mol

R = universal gas constant = 8.314 J/K/mole

T = temperature = 25^0C= (25+273)K=298 K

K_c = equilibrium constant = ?

Putting in the values we get:

-14500=-2.303\times 8.314\times 298\times \log K_c

\log K_c=2.54

K_c=antilog(2.54)=346.7

The equilibrium constant for the reaction at 25 °C is 346.7

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What information do we need to determine the molecular formula of a compound from the empirical formula?
defon

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7 0
3 years ago
Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3). How much sulfur dioxide
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2) Molar ratios

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number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.

4) Use proportions with the molar ratios

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=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.

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