What is the value of PV type of work for the | vedclass

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(D) The reaction is $C_2H_4(g) + HCl(g) \rightarrow C_2H_5Cl(g)$.$1$ mole of $C_2H_4$ reacts with $1$ mole of $HCl$ to produce $1$ mole of $C_2H_5Cl$.

Vedclass · Accepted Answer

$15$

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(D) The reaction is $C_2H_4(g) + HCl(g) ightarrow C_2H_5Cl(g)$.$1$ mole of $C_2H_4$ reacts with $1$ mole of $HCl$ to produce $1$ mole of $C_2H_5Cl$.Given: $V(C_2H_4) = 200 \ mL$,$V(HCl) = 150 \ mL$.Since $HCl$ is the limiting reagent,$150 \ mL$ of $HCl$ reacts with $150 \ mL$ of $C_2H_4$ to produce $150 \ mL$ of $C_2H_5Cl$.Initial volume $V_1 = 150 \ mL (C_2H_4) + 150 \ mL (HCl) = 300 \ mL = 0.3 \ dm^3$.Final volume $V_2 = 150 \ mL (C_2H_5Cl) = 0.15 \ dm^3$.Work $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.$W = -1 \ bar 	imes (0.15 \ dm^3 - 0.3 \ dm^3) = -1 \ bar 	imes (-0.15 \ dm^3) = 0.15 \ dm^3 \ bar$.Since $1 \ dm^3 \ bar = 100 \ J$,$W = 0.15 	imes 100 \ J = 15 \ J$.

What is the value of $PV$ type of work for the following reaction at $1 \ bar$ pressure (in $J$)? The reaction is: $C_2H_4(g) + HCl(g) \rightarrow C_2H_5Cl(g)$. Given volumes are $200 \ mL$ of $C_2H_4$ and $150 \ mL$ of $HCl$.

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