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Question Number 51245 by Tawa1 last updated on 25/Dec/18

Answered by afachri last updated on 25/Dec/18

$$\left(2\right)\mathbf{to}\mathbf{evaluate}△H_{\mathrm{Reaction}}\mathbf{with}\mathbf{bond}\mathbf{energy}:$$$$△H_{\mathrm{Reaction}}=△H_{\mathrm{r}}-△H_{\mathrm{p}}$$$$△H_{\mathrm{p}};(\mathrm{N}-\mathrm{N})+4(\mathrm{N}-\mathrm{H})=(+167)+4(385.9)$$$$=+1710.6\mathrm{Kj}/\mathrm{mol}$$$$△H_{\mathrm{r}};2(\mathrm{H}-\mathrm{H})+(\mathrm{N}=\mathrm{N})=2(437.52)+(95.14)$$$$=+970.18\mathrm{Kj}/\mathrm{mol}$$$$$$$$\mathrm{so},△H_{\mathrm{Reaction}}=970.18-1710.6$$$$=-740.42\mathrm{Kj}/\mathrm{mol}\left(\mathit{e}\mathit{x}\mathit{o}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{i}\mathit{c}\right)$$

Commented by Tawa1 last updated on 25/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by afachri last updated on 25/Dec/18

$$(3.\mathrm{i}\mathbf{i})\mathit{a}\mathrm{Fe}{\left(\mathrm{CO}\right)}_5+\mathit{b}\mathrm{NaOH}\to \mathit{c}{\mathrm{Na}}_2\mathrm{Fe}{\left(\mathrm{CO}\right)}_4+\mathit{d}{\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O}$$$$\mathbf{H};b=2$$$$\mathbf{Fe};a=c$$$$\mathrm{So},\mathrm{eqn}\mathrm{of}\mathrm{reaction}\mathrm{will}\mathrm{be}$$$$\mathit{a}\mathrm{Fe}{\left(\mathrm{CO}\right)}_5+2\mathrm{NaOH}\to \mathit{a}{\mathrm{Na}}_2\mathrm{Fe}{\left(\mathrm{CO}\right)}_4+\mathit{d}{\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O}$$$$\mathbf{Na};2=2a+2d\Rightarrow a+d=1$$$$\mathbf{C};5a=4a+d\Rightarrow a-d=0$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$2d=1$$$$d=\frac{1}{2}\mathrm{and}\mathrm{so},a=\frac{1}{2}$$$$c=a=\frac{1}{2};b=2$$$$\frac{1}{2}\mathrm{Fe}{\left(\mathrm{CO}\right)}_5+2\mathrm{NaOH}\to \frac{1}{2}{\mathrm{Na}}_2\mathrm{Fe}{\left(\mathrm{CO}\right)}_4+\frac{1}{2}{\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O}$$$$\mathit{f}\mathit{i}\mathit{n}\mathit{a}\mathit{l}\mathit{r}\mathit{e}\mathit{s}\mathit{u}\mathit{l}\mathit{t}$$$$\mathrm{Fe}{\left(\mathrm{CO}\right)}_5+4\mathrm{NaOH}\to {\mathrm{Na}}_2\mathrm{Fe}{\left(\mathrm{CO}\right)}_4+{\mathrm{Na}}_2{\mathrm{CO}}_3+{2\mathrm{H}}_2\mathrm{O}$$$$$$

Commented by Tawa1 last updated on 25/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by afachri last updated on 25/Dec/18

$$\mathrm{ur}\mathrm{welcome}\mathrm{Sir}$$

Commented by rahul 19 last updated on 25/Dec/18

$$SirAfachridoyouknowchemistry?$$$$ImeanCanUtellmychemistrydoubts?$$

Commented by malwaan last updated on 25/Dec/18

$$\mathrm{great}$$$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Commented by afachri last updated on 25/Dec/18

$${\mathrm{what}}^{\prime }\mathrm{s}\mathrm{that}\mathrm{your}\mathrm{doubts}\mathrm{Mr}\mathrm{Rahul}??\mathrm{if}\mathrm{i}$$$$\mathrm{can}\mathrm{help}\mathrm{you}\stackrel{}{.}$$

Commented by afachri last updated on 25/Dec/18

$$:)))$$

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