Calculate the work done (in J) when 4.5 g of | vedclass

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(A) The balanced chemical equation is: $2H_2O_{2(l)} 	o O_{2(g)} + 2H_2O_{(l)}$.
Molar mass of $H_2O_2 = 34 \ g \cdot mol^{-1}$.
Moles of $H_2O_2 =

Vedclass · Accepted Answer

$-1.63 	imes 10^2$

Vedclass · Answer

(A) The balanced chemical equation is: $2H_2O_{2(l)} 	o O_{2(g)} + 2H_2O_{(l)}$.
Molar mass of $H_2O_2 = 34 \ g \cdot mol^{-1}$.
Moles of $H_2O_2 = \frac{4.5 \ g}{34 \ g \cdot mol^{-1}} \approx 0.13235 \ mol$.
From the stoichiometry,$2 \ mol$ of $H_2O_2$ produces $1 \ mol$ of $O_{2(g)}$.
Therefore,moles of $O_{2(g)}$ produced $(\Delta n_g) = \frac{0.13235}{2} = 0.066175 \ mol$.
Work done $w = -P_{ext} \Delta V = -\Delta n_g RT$.
Using $R = 8.314 \ J \cdot K^{-1} \cdot mol^{-1}$ and $T = 298 \ K$:
$w = -(0.066175 \ mol) 	imes (8.314 \ J \cdot K^{-1} \cdot mol^{-1}) 	imes (298 \ K) \approx -163.9 \ J$.
Rounding to two significant figures,$w = -1.64 	imes 10^2 \ J$.

Condition	Details
$A$. $\Delta H: +, \Delta S: -$	$T$: any $T$,Spontaneity: Non-spontaneous
$B$. $\Delta H: +, \Delta S: +$	$T$: low $T$,Spontaneity: Non-spontaneous
$C$. $\Delta H: -, \Delta S: -$	$T$: low $T$,Spontaneity: Spontaneous
$D$. $\Delta H: -, \Delta S: +$	$T$: any $T$,Spontaneity: Spontaneous

Calculate the work done (in $J$) when $4.5 \ g$ of $H_2O_2$ reacts against a pressure of $1.0 \ atm$ at $25 \ ^oC$. [$2H_2O_{2(l)} \to O_{2(g)} + 2H_2O_{(l)}$]

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