Calculate the enthalpy of formation of carbon | vedclass

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Question

(A) We need to find the enthalpy of formation for the reaction: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \ (I)$ Given equations: $C_{(s)}

Vedclass · Accepted Answer

$-111.1 \ kJ \ mol^{-1}$

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(A) We need to find the enthalpy of formation for the reaction: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \ (I)$ Given equations: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H_1 = -393.3 \ kJ \ mol^{-1} \ (II)$ $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H_2 = -282.2 \ kJ \ mol^{-1} \ (III)$ Subtracting equation $(III)$ from equation $(II)$: $(C_{(s)} + O_{2(g)}) - (CO_{(g)} + \frac{1}{2} O_{2(g)}) \rightarrow CO_{2(g)} - CO_{2(g)}$ $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$ Therefore,$\Delta H_f = \Delta H_1 - \Delta H_2$ $\Delta H_f = -393.3 - (-282.2) = -111.1 \ kJ \ mol^{-1}$

Calculate the enthalpy of formation of carbon monoxide $(CO).$ Given: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -393.3 \ kJ \ mol^{-1}$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -282.2 \ kJ \ mol^{-1}.$

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