Question

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Answer 1

QUESTION:

For the reaction 2NO (g) + O₂ (g) ⟶ 2NO₂ (g). Calculate ΔG at 700 K when enthalpy and entropy (∆H and ∆S) changes are - 113 kJ/mol - 145 J/Kmol respectively.

ANSWER:

• The value of ΔG = -11.5 kJ/mol

GIVEN:

• Reaction is 2NO (g) + O₂ (g) ⟶ 2NO₂ (g).

• ∆H and ∆S changes are - 113 kJ/mol - 145 J/Kmol respectively.

• Temperature = 700 K.

TO FIND:

• The value of ΔG.

EXPLANATION:

$\boxed{\boldsymbol{\large{\green{\Delta G = \Delta H- T \Delta S}}}}$

$\tt \implies \Delta H = -113\ kJmol^{-1}$

$\tt \implies T = 700\ K$

$\tt \implies \Delta S = -145 \ JK^{-1}mol^{-1}$

$\sf \Delta G = (-113 \times {10}^{3}) - 700( -145)$

$\sf \Delta G = (-113 \times {10}^{3}) + 101500$

$\sf \Delta G = (-113 \times {10}^{3}) + (101.5 \times {10}^{3} )$

$\sf \Delta G = (-113 + 101.5) \times {10}^{3}$

$\sf \Delta G = - 11.5 \times {10}^{3} \ Jmol^{-1}$

$\sf \Delta G = - 11.5 \ kJmol^{-1}$

Hence the value of ΔG = - 11.5 kJ/mol.

Answer 2

$\huge\color{purple}{\underline{\underline{Answer\::}}}$

∆ G = -11.5 KJmol ^ -¹

$\huge\color{blue}{\underline{\underline{explaination\::}}}$

Refers to the attachment....