Subjective Type

For the reaction, $$C_{graphite} +\dfrac{1}{2} O_2(g)=CO(g)$$ at $$298 K$$ and 1 atm, $$\Delta H=-26416$$ cal. If the molar volume of graphite is $$0.0053$$ litre, calculate $$\Delta U$$.

Solution

$$G_{graphite}+\dfrac{1}{2} O_2(g)\rightarrow Co(g)$$ at 298 K and 1 atm

$$\Delta H=-26416$$ cal

$$V_{molar}=0.0053$$ lit

$$\Delta H=\Delta E+P(V_f-V_i)$$

$$\Delta E+1((1-\dfrac{1}{2})\times 24.466-0.0053)$$ atm lit

$$\therefore at T=298 K$$ and $$P=1$$atm ($$V_{molar}=24.46$$lit)

$$-26416= \Delta E+296.94$$ cal

$$\Delta E= -26416-296.54= -26712.52$$ cal

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