Calculate Delta H for the following reaction | vedclass

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(C) The change in the number of moles of gaseous species is given by $\Delta n_g = (n_{products, g} - n_{reactants, g})$.For the reaction: $NH_2CN_{(g

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$-741.7 \ kJ \ mol^{-1}$

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(C) The change in the number of moles of gaseous species is given by $\Delta n_g = (n_{products, g} - n_{reactants, g})$.For the reaction: $NH_2CN_{(g)} + \frac{3}{2} O_{2_{(g)}} \longrightarrow N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$$\Delta n_g = (1 + 1) - (1 + 1.5) = 2 - 2.5 = -0.5 \ mol$.Given $\Delta U = -740.5 \ kJ$,$T = 25 + 273 = 298 \ K$,and $R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.Using the relation $\Delta H = \Delta U + \Delta n_g RT$:$\Delta H = -740.5 \ kJ + (-0.5 \ mol) 	imes (8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) 	imes 298 \ K$.$\Delta H = -740.5 \ kJ - 1.2388 \ kJ = -741.7388 \ kJ \ mol^{-1}$.Rounding to one decimal place,we get $\Delta H \approx -741.7 \ kJ \ mol^{-1}$.

Calculate $\Delta H$ for the following reaction at $25^{\circ} C$.
$NH_2CN_{(g)} + \frac{3}{2} O_{2_{(g)}} \longrightarrow N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
$(\Delta U = -740.5 \ kJ, R = 8.314 \ J \ K^{-1} \ mol^{-1})$

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Calculate $\Delta H$ for the following reaction at $25^{\circ} C$.$NH_2CN_{(g)} + \frac{3}{2} O_{2_{(g)}} \longrightarrow N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$$(\Delta U = -740.5 \ kJ, R = 8.314 \ J \ K^{-1} \ mol^{-1})$