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gtnhenbr [62]
3 years ago
14

The product(s) of a heated reaction of N-ethyl-N-methylpentanamide with sodium hydroxide would be __________. The product(s) of

a heated reaction of N-ethyl-N-methylpentanamide with sodium hydroxide would be __________. pentanoic acid and ethylmethylamine sodium propanoate and pentylamine sodium pentanoate and ethylmeth
Chemistry
1 answer:
olga2289 [7]3 years ago
6 0

Answer:

sodium pentanoate and ethylmethylamine

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Answer: -105 kJ

Explanation:-

The balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]

\Delta H=[(n_{N_2}\times B.E_{N_2})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{NH_3}\times B.E_{NH_3})]

\Delta H=[(n_{N_2}\times B.E_{N\equiv N})+(n_{H_2}\times B.E_{H-H}) ]-[(n_{NH_3}\times 3\times B.E_{N-H})]

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n = number of moles

Now put all the given values in this expression, we get

\Delta H=[(1\times 945)+(3\times 432)]-[(2\times 3\times 391)]

Delta H=-105kJ

Therefore, the enthalpy change for this reaction is, -105 kJ

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