The enthalpy of combustion of benzene from th | vedclass

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(D) The combustion of benzene is represented by the equation:$C_6H_{6(l)} + \frac{15}{2}O_{2(g)} 	o 6CO_{2(g)} + 3H_2O_{(l)}$To find the enthalpy of

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$-3264.6$

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(D) The combustion of benzene is represented by the equation:$C_6H_{6(l)} + \frac{15}{2}O_{2(g)} 	o 6CO_{2(g)} + 3H_2O_{(l)}$To find the enthalpy of combustion,we manipulate the given equations:$1$. Reverse equation $(i)$:$C_6H_{6(l)} 	o 6C_{(s)} + 3H_{2(g)} ; \Delta H = -45.9 \ kJ$$2$. Multiply equation $(ii)$ by $3$:$3H_{2(g)} + \frac{3}{2}O_{2(g)} 	o 3H_2O_{(l)} ; \Delta H = 3 	imes (-285.9) = -857.7 \ kJ$$3$. Multiply equation $(iii)$ by $6$:$6C_{(s)} + 6O_{2(g)} 	o 6CO_{2(g)} ; \Delta H = 6 	imes (-393.5) = -2361.0 \ kJ$Adding these equations gives:$\Delta H_{combustion} = -45.9 - 857.7 - 2361.0 = -3264.6 \ kJ$The enthalpy of combustion of benzene is $-3264.6 \ kJ$.

The enthalpy of combustion of benzene from the following data will be
$(i) \ 6C_{(s)} + 3H_{2(g)} \to C_6H_{6(l)} ; \Delta H = +45.9 \ kJ$
$(ii) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)} ; \Delta H = -285.9 \ kJ$
$(iii) \ C_{(s)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -393.5 \ kJ$
.....$kJ$

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The enthalpy of combustion of benzene from the following data will be$(i) \ 6C_{(s)} + 3H_{2(g)} \to C_6H_{6(l)} ; \Delta H = +45.9 \ kJ$$(ii) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)} ; \Delta H = -285.9 \ kJ$$(iii) \ C_{(s)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -393.5 \ kJ$.....$kJ$