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Anna35 [415]
3 years ago
8

A student determines the value of the equilibrium constant to be 1.5297 x 107 for the following reaction: HBr(g) + 1/2 Cl2(g) --

> HCl(g) +1/2 Br2(g) Based on this value of Keq, calculate the Gibbs free energy change for the reaction of 2.37 moles of HBr(g) at standard conditions at 298 K.
Chemistry
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

\Delta G=-97.14kJ

Explanation:

Hello,

In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

\Delta G=-RTln(K)

Hence, we compute it as required:

\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol

And for 2.37 moles of hydrogen bromide, we obtain:

\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ

Best regards.

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At a certain temperature, 3.67 mol SO 2 and 1.83 mol O 2 are placed in a container. 2 SO 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 SO 3 ( g
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Answer : The number of moles of SO_2 and O_2 at equilibrium is, 1.75 mol and 0.87 mol respectively.

Explanation :

The given chemical reaction is:

                       2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

Initial mol.       3.67         1.83          0

At eqm.        (3.67-2x)   (1.83-x)       2x

As we are given:

Number of moles of SO_3 at equilibrium = 1.92

That means,

2x = 1.92

x = 0.96 mol

Number of moles of SO_2 at equilibrium = (3.67-2x) = [3.67-2(0.96)] = 1.75 mol

Number of moles of O_2 at equilibrium = (1.83-x) = (1.83-0.96) = 0.87 mol

Thus, the number of moles of SO_2 and O_2 at equilibrium is, 1.75 mol and 0.87 mol respectively.

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