32g of CaC₂ react at 27°C with excess of water to produce acetylene gas. What is work done?

a)-2400 Cal b)-300 Cal c)-300 J d)-2400 L-atm

Please explain with detailed explanation

Write down the balanced reaction.

CaC₂ + 2 H₂O (excess) → C₂H₂ + Ca(OH)₂

Molar Mass of CaC₂ = 40 + (12*2) = 40 + 24 = 64 g/mol

No. of Mole of CaC₂ = 32/64 = 0.5 mole

Now, Work Done = Δn_gRT,

where, Δn_g is difference in no. of (gaseous) mole of reactant and product;

R is the Universal Gas Constant;

T is the absolute temperature.

Thus, W = (-0.5)(8.314)(300) = 1247.1 J

or W = (-0.5)(0.0821)(300) = 12.315 L-atm

or W = (-0.5)(1.987)(300) = 298.05 cal ≈ 300 cal

Hence option (b) is correct.