Calculate the work done in the following reac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The work done in a chemical reaction is given by the formula $W = -\Delta n_g RT$.First,calculate the change in the number of moles of gaseous spe

Vedclass · Accepted Answer

$4988.4$

Vedclass · Answer

(A) The work done in a chemical reaction is given by the formula $W = -\Delta n_g RT$.First,calculate the change in the number of moles of gaseous species,$\Delta n_g = n_{p(g)} - n_{r(g)}$.For the reaction $4 SO_{2(g)} + 2 O_{2(g)} ightarrow 4 SO_{3(g)}$,the number of moles of gaseous products is $4$ and the number of moles of gaseous reactants is $4 + 2 = 6$.So,$\Delta n_g = 4 - 6 = -2 \ mol$.The temperature $T = 27^{\circ} C = 27 + 273 = 300 \ K$.The gas constant $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.Now,substitute the values into the formula: $W = -(-2 \ mol) 	imes (8.314 \ J \ K^{-1} \ mol^{-1}) 	imes (300 \ K)$.$W = 2 	imes 8.314 	imes 300 = 4988.4 \ J$.

Calculate the work done in the following reaction at $27^{\circ} C$.
$4 SO_{2(g)} + 2 O_{2(g)} \rightarrow 4 SO_{3(g)}$
$(R = 8.314 \ J \ K^{-1} \ mol^{-1})$ (in $J$)

Similar Questions

What is the relationship between Gibbs free energy change,enthalpy change,and entropy change?

The maximum work done in expanding $16 \ g$ of oxygen at $300 \ K$ from an initial volume of $5 \ dm^3$ to a final volume of $25 \ dm^3$ isothermally and reversibly is $(\log_{10} 5 = 0.699)$.

Calculate the work done if $1 \ mole$ of a certain gas is compressed isothermally and reversibly at $300 \ K$ from an initial pressure $x \ bar$ to a final pressure $2x \ bar$ $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$. (in $kJ$)

$A$ reaction occurs spontaneously if

At a temperature of $25^\circ C$,if $\Delta H < T \Delta S$,then $\Delta G$ = ?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Calculate the work done in the following reaction at $27^{\circ} C$.$4 SO_{2(g)} + 2 O_{2(g)} \rightarrow 4 SO_{3(g)}$$(R = 8.314 \ J \ K^{-1} \ mol^{-1})$ (in $J$)