Nitroglycerine (MW = 227.1) detonates accordi | vedclass

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Question

(B) The balanced chemical equation is:$2 C_{3}H_{5}(NO_{3})_{3(l)} \longrightarrow 3 N_{2(g)} + \frac{1}{2} O_{2(g)} + 6 CO_{2(g)} + 5 H_{2}O_{(g)}$$\

Vedclass · Accepted Answer

$-62.5$

Vedclass · Answer

(B) The balanced chemical equation is:$2 C_{3}H_{5}(NO_{3})_{3(l)} \longrightarrow 3 N_{2(g)} + \frac{1}{2} O_{2(g)} + 6 CO_{2(g)} + 5 H_{2}O_{(g)}$$\Delta H_{	ext{reaction}}^{\circ} = \Sigma \Delta H_{f(P)}^{\circ} - \Sigma \Delta H_{f(R)}^{\circ}$$\Delta H_{	ext{reaction}}^{\circ} = [3(0) + 0.5(0) + 6(-393.5) + 5(-241.8)] - [2(-364)]$$= [-2361 - 1209] - [-728]$$= -3570 + 728 = -2842 \, kJ$ for $2 \, moles$ of nitroglycerine.Enthalpy change for $1 \, mole$ of nitroglycerine $= \frac{-2842}{2} = -1421 \, kJ/mol$.Given $MW = 227.1 \, g/mol$,the enthalpy change for $10 \, g$ is:$\Delta H = \frac{-1421 \, kJ/mol}{227.1 \, g/mol} 	imes 10 \, g \approx -62.57 \, kJ$.Rounding to the nearest option,the answer is $-62.5 \, kJ$.

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