The heat of transition (Delta H_t) of graphit | vedclass

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(B) We are given the following thermochemical equations:$(1) \ C(	ext{graphite}) + O_{2(g)} 	o CO_{2(g)}; \Delta H_1 = x \ kJ \ mol^{-1}$$(2) \ C(

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$(x - y) \ kJ \ mol^{-1}$

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(B) We are given the following thermochemical equations:$(1) \ C(	ext{graphite}) + O_{2(g)} 	o CO_{2(g)}; \Delta H_1 = x \ kJ \ mol^{-1}$$(2) \ C(	ext{diamond}) + O_{2(g)} 	o CO_{2(g)}; \Delta H_2 = y \ kJ \ mol^{-1}$To find the heat of transition for $C(	ext{graphite}) 	o C(	ext{diamond})$,we subtract equation $(2)$ from equation $(1)$:$C(	ext{graphite}) - C(	ext{diamond}) = x - y$$C(	ext{graphite}) 	o C(	ext{diamond}); \Delta H_t = (x - y) \ kJ \ mol^{-1}$

The heat of transition $(\Delta H_t)$ of graphite into diamond would be,where
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = x \ kJ \ mol^{-1}$
$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = y \ kJ \ mol^{-1}$

Similar Questions

Consider the reactions:
$C_{(s)} + 2H_{2(g)} \to CH_{4(g)}, \Delta H = -x \ kcal$
$C_{(g)} + 4H_{(g)} \to CH_{4(g)}, \Delta H = -x_1 \ kcal$
$CH_{4(g)} \to CH_{3(g)} + H_{(g)}, \Delta H = +y \ kcal$
The bond energy of $C-H$ bond is:

The bond energy (in $kcal \ mol^{-1}$) of a $C-C$ single bond is approximately:

Which of the following reactions defines $\Delta H_f^o$?

Which of the following reactions is endothermic?

Given the thermochemical reactions:
$C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta H = -110.5 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta H = -283.2 \ kJ$
Calculate the heat of reaction for $C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}$ in $kJ$.

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The heat of transition $(\Delta H_t)$ of graphite into diamond would be,where$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = x \ kJ \ mol^{-1}$$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = y \ kJ \ mol^{-1}$