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3 years ago

The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4

1 answer:

5 0

In NH3 , let oxidation number of N be x

x + (+1)3 = 0

x = -3

In HNO3 , let oxidation number of N be x

1 + x + (-2)3 = 0

x = +5

In NO2 , let oxidation number of N be x

x + (-2)2 = 0

x = +4

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3 years ago

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3 years ago

Using the bond energy data from your text (or the internet), determine (show calculations for) the approximate enthalpy change ,

aleksley [76]

Answer:

∆H= 438 KJ/mol

Explanation:

First, we have to find the energy bond values for each compound:

-) Cl-Cl = 243 KJ/mol

-) F-F = 159 KJ/mol

-) F-Cl = 193 KJ/mol

If we check the reaction we can calculate the number of bonds:

$Cl_2_(_g_)~+~3F_2_(_g_)~->~2ClF_3_(_g_)$

In total we will have:

-) Cl-Cl = 1

-) F-F = 3

-) F-Cl = 6

With this in mind. we can calculate the total energy for each bond:

-) Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol

-) F-F = (3*159 KJ/mol) = 477 KJ/mol

-) F-Cl = (6*193 KJ/mol) = 1158 KJ/mol

Now, we can calculate the total energy of the products and the reagents:

Reagents = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol

Products = 1158 KJ/mol

Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:

∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol

I hope it helps!

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3 years ago