When the following reaction was carried out in a bomb calorimeter, Δ U is found to be - text740 text.0 textkJ/mol of ( textNH) text2 textCN(s) (s) at text300 textK . NH2CN(. s .)+ (3/2)O2 (g) arrow N2 (g)+ CO2 (g)+ H2O(. l .) Calculate textÎH text300 textK for the reaction.

Question

When the following reaction was carried out in a bomb calorimeter, Δ U is found to be - text740 text.0 textkJ/mol of ( textNH) text2 textCN(s) (s) at text300 textK . NH2CN(. s .)+ (3/2)O2 (g) arrow N2 (g)+ CO2 (g)+ H2O(. l .) Calculate textÎH text300 textK for the reaction. (A) -738.75 kJ (B) +738.75 kJ (C) -824.75 kJ (D) -919.57 kJ

Tardigrade Admin · Accepted Answer

textΔH text= textΔU text+ textΔn textg textRT textΔH = - 740000 + 0.5 × 8.3 × 300 = - 738755 textJ

Q. When the following reaction was carried out in a bomb calorimeter, $Δ U$ is found to be $- 740 .0 kJ/mol$ of $(NH)_{2} CN(s)$ (s) at $300 K$ .

$N H_{2} C N_{(s)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + C O_{2 (g)} + H_{2} O_{(l)}$

Calculate $ΔH_{300 K}$ for the reaction.

-738.75 kJ

+738.75 kJ

-824.75 kJ

-919.57 kJ

$ΔH = ΔU + Δn_{g} RT$

$ΔH = - 740000 + 0.5 \times 8.3 \times 300 = - 738755 J$

Q. When the following reaction was carried out in a bomb calorimeter, ΔU is found to be −740.0kJ/mol of (NH)2​CN(s) (s) at 300K . NH2​CN(s)​+23​O2(g)​→N2(g)​+CO2(g)​+H2​O(l)​ Calculate ΔH300K​ for the reaction.