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NemiM [27]
4 years ago
13

The following reaction is found to be at equilibrium at 25 celcius: 2SO3--->O2 + 2SO2 + 198kJ/mol. If the value for Kc at 25

celcius is 8.1, are the products or reactants favored and how much?
A.) Reactants are strongly favored
B.) Reactants are weakly favored
C.) Products are strongly favored
D.) Products are weakly favored
Chemistry
1 answer:
kondaur [170]4 years ago
4 0

Answer:

D.) Products are weakly favored

Explanation:

For the reaction:

2SO₃ ⇄ O₂ + 2SO₂ + 198kJ/mol

The kc is defined as:

kc = [O₂] [SO₂]² / [SO₃]²

As the kc is 8,1:

8,1 [SO₃]² =  [O₂] [SO₂]²

The products are favored 8,1 times. This is a weakly favored because the usual kc are in the order of 1x10⁴. Thus, right answer is:

D.) Products are weakly favored

I hope it helps!

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By titration, it is found that 31.7 mL of 0.145 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentrati
arlik [135]

Answer:

0.184 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above, the following data were obtained:

Mole ratio of the acid, HCl (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, the data obtained from the question. This includes:

Volume of the base, NaOH (Vb) = 31.7 mL Molarity of the base, NaOH (Mb) = 0.145 M

Volume of the acid, HCl (Va) = 25.0 mL

Molarity of the acid, HCl (Ma) =?

Finally, we shall determine the molarity of the acid (HCl) as shown below:

MaVa /MbVb = nA/nB

Ma × 25 / 0.145 × 31.7 = 1

Cross multiply

Ma × 25 = 0.145 × 31.7

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Divide both side by 25

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Therefore, the molarity of the acid (HCl) is 0.184 M

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