When 1.0 g of solid oxalic acid (H_2C_2O_4) | vedclass

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(D) The combustion reaction is: $H_2C_2O_{4(s)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(\ell)} + 2CO_{2(g)}$The change in the number of gaseous mole

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$-241.96 \ kJ/mol$

Vedclass · Answer

(D) The combustion reaction is: $H_2C_2O_{4(s)} + \frac{1}{2} O_{2(g)} ightarrow H_2O_{(\ell)} + 2CO_{2(g)}$The change in the number of gaseous moles is $\Delta n_g = 2 - 0.5 = 1.5$.The internal energy of combustion $(\Delta U_c)$ is calculated as: $\Delta U_c = -\frac{q}{n} = -\frac{C 	imes \Delta T}{n}$.Given $C = 8.75 \ kJ/K$,$\Delta T = 0.312 \ K$,and molar mass of $H_2C_2O_4 = 90 \ g/mol$,the moles $n = \frac{1.0 \ g}{90 \ g/mol} = \frac{1}{90} \ mol$.$\Delta U_c = -\frac{8.75 	imes 0.312}{1/90} = -245.7 \ kJ/mol$.Using the relation $\Delta H = \Delta U + \Delta n_g RT$ at $T = 300 \ K$:$\Delta H = -245.7 + (1.5 	imes 8.314 	imes 10^{-3} 	imes 300) = -245.7 + 3.7413 = -241.9587 \ kJ/mol \approx -241.96 \ kJ/mol$.

Column $I$	Column $II$
$(a)$ Adiabatic process	$(1)$ Heat
$(b)$ Isolated system	$(2)$ Constant volume
$(c)$ Isothermal change	$(3)$ First law of thermodynamics
$(d)$ Path function	$(4)$ No exchange of matter and energy
$(e)$ State function	$(5)$ No heat exchange
$(f)$ $\Delta U = q$	$(6)$ Constant temperature
$(g)$ Law of conservation of energy	$(7)$ Internal energy
$(h)$ Reversible process	$(8)$ $p_{ext} = 0$
$(i)$ Free expansion	$(9)$ Constant pressure
$(j)$ $\Delta H = q$	$(10)$ Infinitely slow process involving equilibrium states
$(k)$ Intensive property	$(11)$ Entropy
$(l)$ Extensive property	$(12)$ Pressure
	$(13)$ Specific heat

When $1.0 \ g$ of solid oxalic acid $(H_2C_2O_4)$ is burned in a bomb calorimeter whose heat capacity is $8.75 \ kJ/K$,the temperature increases by $0.312 \ K$. The enthalpy of combustion of oxalic acid at $27 \ ^oC$ is

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