Question

The work done in an open vessel at 300 k, when 56g iron reacts with dilute hcl

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Temperature given (T) = 300K.
• Given Weight of Iron (w) = 56 grams.

$\huge{\bold{\underline{Explanation:-}}}$

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The reaction will occur in a open vessel, Therefore,

$\large{\boxed{\tt W = - p_{ext} \Delta V}}$

Here,

• "W" = Work done.
• ${\tt p_{ext}}$ = External Pressure applied.
• ${\tt \Delta V}$ = Change in Volume.

Now,

$\large{\boxed{\tt \Delta V = V_{final} - V_{initial}}}$

$\large{\tt \leadsto \Delta V = V_{final} - V_{initial}}$

$\large{\tt \leadsto \Delta V = V_{final} - V_{initial} \approx V_{final}}$

Here The initial Volume has a very low concentration as it is solid ,

Therefore Can be neglected,

$\large{\underline{\tt \leadsto \Delta V \approx V_{final}}}$

Now, Applying Ideal gas equation,

$\large{\boxed{\tt V = \dfrac{nRT}{P}}}$

Substituting the values,

$\large{\underline{\tt \leadsto V_{final} = \dfrac{nRT}{P_{ext}}}}$

Applying in the work done formula,

$\large{\boxed{\tt W = - p_{ext} \Delta V}}$

$\large{\tt \leadsto W = - P_{ext} \times V_{final} \: \: \: \: [\Delta V \approx V_{final}]}$

Substituting the Volume,

$\large{\tt \leadsto W = - P_{ext} \times \dfrac{nRT}{P_{ext}}}$

$\large{\tt \leadsto W = - \cancel{P_{ext}} \times \dfrac{nRT}{\cancel{P_{ext}}}}$

$\large{\tt W = - nRT \: -----(1) }$

$\large{\boxed{\tt W = - nRT}}$

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Now, Writing the Chemical Reaction,

$\large{\bold{\tt Fe_{(s)} + 2HCl_{(aq)} \xrightarrow{} FeCl_{2_{(aq)}}+ H_{2_{(g)}}}}$

Finding The Number of moles,

$\large{\boxed{\tt No. \: of \: moles = \dfrac{Given \: weight}{Molar \: Mass}}}$

Substituting the values,

$\large{\tt \leadsto n = \dfrac{56}{56}}$

$\large{\tt \leadsto n = \dfrac{\cancel{56}}{\cancel{56}}}$

$\large{\underline{\tt \leadsto n = 1 \: mole}}$

Now, Substituting it in the equation (1).

$\large{\tt W = - nRT}$

• T = 300K,
• R = 2 calories,
• n = 1 mole,

$\large{\tt \leadsto W = - 1 \times 2 \times 300}$

$\large{\tt \leadsto W = - 2 \times 300}$

$\huge{\boxed{\boxed{\tt W = - 600 \: calories}}}$

Thus, The system Does 600 calories work on the atmosphere.

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