Calculate the standard enthalpy of formation of $$CH_3OH$$(l) from the following data:

$$CH_3OH(l)$$ + $$\dfrac{3}{2}$$ $$O_2$$(g)→ $$CO_2$$(g) + $$2H_2O$$(l);$$\Delta _{r}H^{\Theta }= -726 kJ mol^{-1}$$

C(g) + $$O_2$$(g)→ $$CO_2$$(g); $$\Delta _{c}H^{\Theta }= -393 kJ mol^{-1}$$

$$H_2$$(g) + $$\dfrac{1}{2}$$$$O_2$$(g)→ $$H_2O$$(l); $$\Delta _{f}H^{\Theta } = -286 kJ mol^{-1}$$

Asked by Pragya Singh | 1 year ago |  66

Solution :-

C(s) + 2H2O(g) + ($$\dfrac{1}{2}$$)O2(g) → CH3OH(l) ……………(i)

CH3OH(l) can be obtained as follows,

$$\Delta _{f}H_{\Theta }[CH3OH_{(l)}] = \Delta _{c}H_{\Theta }$$

$$2\Delta _{f}H_{\Theta }– \Delta _{r}H_{\Theta }$$

$$(-393 kJ mol^{-1}) +2(-286kJ mol^{-1})$$

$$– (-726kJ mol^{-1})$$

=$$(-393 – 572 + 726) kJ mol^{-1}$$

$$-239kJ mol^{-1}$$

Thus,

$$\Delta _{f}H_{\Theta }​ [CH3OH_{(l)}]$$

$$= -239kJ mol^{-1}$$

Answered by Pragya Singh | 1 year ago

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