State Hess's Law of Constant Heat Summation.
(N/A) Hess's Law of Constant Heat Summation states that the total enthalpy change for a chemical reaction is the same,whether the reaction takes place in one step or in several steps. It is the algebraic sum of the enthalpy changes of the individual steps of the reaction.
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The heat of transition $(\Delta H_t)$ of graphite into diamond would be,where
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = x \ kJ \ mol^{-1}$
$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = y \ kJ \ mol^{-1}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = x \ kJ \ mol^{-1}$
$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = y \ kJ \ mol^{-1}$
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View SolutionIf the enthalpies of atomization of methane and ethane are respectively $320 \ cal$ and $360 \ cal$,then the bond energy of the $C-C$ bond is......$cal$.
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View SolutionThe following thermochemical reactions are given:
$M + \frac{1}{2} O_2 \to MO + 351.4 \ kJ$ $(i)$
$X + \frac{1}{2} O_2 \to XO + 90.8 \ kJ$ $(ii)$
It follows that the heat of reaction for the following process $M + XO \rightleftharpoons MO + X$ is given by $... \ kJ$.
$M + \frac{1}{2} O_2 \to MO + 351.4 \ kJ$ $(i)$
$X + \frac{1}{2} O_2 \to XO + 90.8 \ kJ$ $(ii)$
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View SolutionEnthalpy change for the reaction,$\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \to HCl(g)$,is called:
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View SolutionIf the value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$,then the formation of one mole of water from $H_{(g)}$ and $O_{(g)}$ is associated with:
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