The ratio of heats liberated at 298 K from t | vedclass

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(A) One $kg$ of coke contains $n = \frac{1000}{12} = 83.33 \ mol$ of carbon. Heat liberated by burning $1 \ kg$ of coke: $C(s) + O_{2}(g) \rightarrow

Vedclass · Accepted Answer

$0.69$

Vedclass · Answer

(A) One $kg$ of coke contains $n = \frac{1000}{12} = 83.33 \ mol$ of carbon. Heat liberated by burning $1 \ kg$ of coke: $C(s) + O_{2}(g) ightarrow CO_{2}(g) ; \Delta H_{1} = 83.33 	imes 393.5 \ kJ$. Water gas is produced by: $C(s) + H_{2}O(g) ightarrow CO(g) + H_{2}(g)$. Thus,$1 \ kg$ of coke produces $83.33 \ mol$ of $CO$ and $83.33 \ mol$ of $H_{2}$. Heat liberated by burning this water gas: $CO(g) + \frac{1}{2}O_{2}(g) ightarrow CO_{2}(g) ; \Delta H_{CO} = 83.33 	imes 283.5 \ kJ$. $H_{2}(g) + \frac{1}{2}O_{2}(g) ightarrow H_{2}O(l) ; \Delta H_{H_{2}} = 83.33 	imes 285.5 \ kJ$. Total heat $\Delta H_{2} = 83.33 	imes (283.5 + 285.5) = 83.33 	imes 569 \ kJ$. Ratio $= \frac{\Delta H_{1}}{\Delta H_{2}} = \frac{393.5}{569} \approx 0.69 : 1$.

Process	Entropy Change
$(a)$ Liquid to vapor conversion	$(1)$ $\Delta S = 0$
$(b)$ Process not spontaneous at any temperature	$(2)$ $\Delta S = (+)$
$(c)$ Reversible expansion of an ideal gas	$(3)$ $\Delta S = (-)$

Similar Questions

$A$ solution of $500 \ mL$ of $0.2 \ M \ KOH$ and $500 \ mL$ of $0.2 \ M \ HCl$ is mixed and stirred; the rise in temperature is $T_1$. The experiment is repeated using $250 \ mL$ each of solution,the temperature raised is $T_2$. Which of the following is true?

Match the following processes with their corresponding entropy changes:

Process Entropy Change

$(a)$ Liquid to vapor conversion $(1)$ $\Delta S = 0$

$(b)$ Process not spontaneous at any temperature $(2)$ $\Delta S = (+)$

$(c)$ Reversible expansion of an ideal gas $(3)$ $\Delta S = (-)$

For the reaction
$A_{(\ell)} \rightarrow 2 B_{(g)}$
$\Delta U = 2.1 \; kcal, \Delta S = 20 \; cal \; K^{-1} \; mol^{-1}$ at $300 \; K$
Hence $\Delta G$ in $kcal \; mol^{-1}$ is

Benzene burns according to the following equation at $300 \ K$ $(R = 8.314 \ J \ mol^{-1} K^{-1})$: $2 C_6H_{6(l)} + 15 O_{2(g)} \to 12 CO_{2(g)} + 6 H_2O_{(l)}$,$\Delta H^o = -6542 \ kJ/mol$. What is the $\Delta E^o$ for the combustion of $1.5 \ mol$ of benzene in $kJ$?

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