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Question-51245

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Question Number 51245 by Tawa1 last updated on 25/Dec/18
Answered by afachri last updated on 25/Dec/18
(2) to evaluate △H_(Reaction) with bond energy : △H_(Reaction) = △H_r − △H_p △H_p ; ( N−N) + 4(N−H) = (+167) + 4(385.9) = +1710.6 Kj/mol △H_r ; 2(H−H) + (N=N) = 2(437.52) + (95.14) = +970.18 Kj/mol so, △H_(Reaction) = 970.18 − 1710.6 = −740.42 Kj/mol (exotermic)
$$\left(\mathrm{2}\right)\:\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{evaluate}}\:\bigtriangleup{H}_{\mathrm{Reaction}} \:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{bond}}\:\boldsymbol{\mathrm{energy}}\:: \\ $$$$\bigtriangleup{H}_{\mathrm{Reaction}} \:=\:\bigtriangleup{H}_{\mathrm{r}} \:−\:\bigtriangleup{H}_{\mathrm{p}} \\ $$$$\bigtriangleup{H}_{\mathrm{p}} \:;\:\left(\:\mathrm{N}−\mathrm{N}\right)\:+\:\mathrm{4}\left(\mathrm{N}−\mathrm{H}\right)\:=\:\left(+\mathrm{167}\right)\:\:+\:\:\mathrm{4}\left(\mathrm{385}.\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:+\mathrm{1710}.\mathrm{6}\:\:\:\mathrm{Kj}/\mathrm{mol} \\ $$$$\bigtriangleup{H}_{\mathrm{r}} \:\:;\:\:\mathrm{2}\left(\mathrm{H}−\mathrm{H}\right)\:+\:\left(\mathrm{N}=\mathrm{N}\right)\:=\:\mathrm{2}\left(\mathrm{437}.\mathrm{52}\right)\:+\:\left(\mathrm{95}.\mathrm{14}\right)\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:+\mathrm{970}.\mathrm{18}\:\:\mathrm{Kj}/\mathrm{mol} \\ $$$$ \\ $$$$\mathrm{so},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\bigtriangleup{H}_{\mathrm{Reaction}} \:=\:\mathrm{970}.\mathrm{18}\:−\:\mathrm{1710}.\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\mathrm{740}.\mathrm{42}\:\:\:\mathrm{Kj}/\mathrm{mol}\:\:\:\left(\boldsymbol{{exotermic}}\right) \\ $$
Commented by Tawa1 last updated on 25/Dec/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by afachri last updated on 25/Dec/18
(3.ii) aFe(CO)_5 + bNaOH → cNa_2 Fe(CO)_4 + d Na_2 CO_3 + H_2 O H ; b = 2 Fe ; a = c So, eqn of reaction will be aFe(CO)_5 + 2NaOH → aNa_2 Fe(CO)_4 + d Na_2 CO_3 + H_2 O Na ; 2 = 2a + 2d ⇒ a + d = 1 C ; 5a = 4a + d ⇒ a − d = 0 _________ _ 2d = 1 d = (1/2) and so , a = (1/2) c = a = (1/2) ; b = 2 (1/2)Fe(CO)_5 + 2NaOH → (1/2)Na_2 Fe(CO)_4 + (1/2) Na_2 CO_3 + H_2 O final result Fe(CO)_5 + 4NaOH → Na_2 Fe(CO)_4 + Na_2 CO_3 + 2H_2 O
$$\left(\mathrm{3}.\mathrm{i}\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{{a}}\mathrm{Fe}\left(\mathrm{CO}\right)_{\mathrm{5}} \:+\:\boldsymbol{{b}}\mathrm{NaOH}\:\:\:\rightarrow\:\:\:\boldsymbol{{c}}\mathrm{Na}_{\mathrm{2}} \mathrm{Fe}\left(\mathrm{CO}\right)_{\mathrm{4}} \:+\:\boldsymbol{{d}}\:\mathrm{Na}_{\mathrm{2}} \mathrm{CO}_{\mathrm{3}} \:+\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$$$\boldsymbol{\mathrm{H}}\:\:\:;\:\:\:{b}\:=\:\mathrm{2} \\ $$$$\boldsymbol{\mathrm{Fe}}\:;\:\:{a}\:=\:{c} \\ $$$$\mathrm{So},\:\mathrm{eqn}\:\mathrm{of}\:\mathrm{reaction}\:\mathrm{will}\:\mathrm{be} \\ $$$$\:\boldsymbol{{a}}\mathrm{Fe}\left(\mathrm{CO}\right)_{\mathrm{5}} \:+\:\mathrm{2NaOH}\:\:\:\rightarrow\:\:\:\boldsymbol{{a}}\mathrm{Na}_{\mathrm{2}} \mathrm{Fe}\left(\mathrm{CO}\right)_{\mathrm{4}} \:+\:\boldsymbol{{d}}\:\mathrm{Na}_{\mathrm{2}} \mathrm{CO}_{\mathrm{3}} \:+\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$$$\boldsymbol{\mathrm{Na}}\:\:\:;\:\:\:\:\:\:\mathrm{2}\:\:\:\:=\:\mathrm{2}{a}\:+\:\mathrm{2}{d}\:\:\:\:\:\Rightarrow\:\:{a}\:+\:{d}\:\:\:\:\:\:\:=\:\mathrm{1} \\ $$$$\boldsymbol{\mathrm{C}}\:\:\:\:\:\:;\:\:\:\mathrm{5}{a}\:\:\:\:=\:\mathrm{4}{a}\:+\:\:{d}\:\:\:\:\:\Rightarrow\:\:\:{a}\:−\:\:{d}\:\:\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\_\_\_\_\_\_\_\_\_\:\:\_ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{d}\:\:\:\:=\:\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{and}\:\mathrm{so}\:,\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}\:=\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:;\:\:\:\:{b}\:\:=\:\mathrm{2} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Fe}\left(\mathrm{CO}\right)_{\mathrm{5}} \:+\:\mathrm{2NaOH}\:\:\:\rightarrow\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Na}_{\mathrm{2}} \mathrm{Fe}\left(\mathrm{CO}\right)_{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{Na}_{\mathrm{2}} \mathrm{CO}_{\mathrm{3}} \:+\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$$$\boldsymbol{{final}}\:\boldsymbol{{result}} \\ $$$$\:\mathrm{Fe}\left(\mathrm{CO}\right)_{\mathrm{5}} \:+\:\mathrm{4NaOH}\:\:\:\rightarrow\:\:\:\mathrm{Na}_{\mathrm{2}} \mathrm{Fe}\left(\mathrm{CO}\right)_{\mathrm{4}} \:+\:\mathrm{Na}_{\mathrm{2}} \mathrm{CO}_{\mathrm{3}} \:+\:\:\mathrm{2H}_{\mathrm{2}} \mathrm{O} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 25/Dec/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by afachri last updated on 25/Dec/18
ur welcome Sir
$$\mathrm{ur}\:\mathrm{welcome}\:\mathrm{Sir} \\ $$
Commented by rahul 19 last updated on 25/Dec/18
Sir Afachri do you know chemistry? I mean Can U tell my chemistry doubts?
$${Sir}\:{Afachri}\:{do}\:{you}\:{know}\:{chemistry}? \\ $$$${I}\:{mean}\:{Can}\:{U}\:{tell}\:{my}\:{chemistry}\:{doubts}? \\ $$
Commented by malwaan last updated on 25/Dec/18
great thank you so much
$$\mathrm{great} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by afachri last updated on 25/Dec/18
what′s that your doubts Mr Rahul ?? if i can help you.^
$$\mathrm{what}'\mathrm{s}\:\mathrm{that}\:\mathrm{your}\:\mathrm{doubts}\:\:\mathrm{Mr}\:\mathrm{Rahul}\:??\:\mathrm{if}\:\mathrm{i} \\ $$$$\mathrm{can}\:\mathrm{help}\:\mathrm{you}\overset{} {.} \\ $$
Commented by afachri last updated on 25/Dec/18
:)))
$$\left.:\left.\right)\left.\right)\right) \\ $$

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