When 0.485 g of compound x is burned completely in a bomb calorimeter containing 3000 g of water, a temperature rise of 0.285°c is observed. what is δu of the reaction for the combustion of compound x? the hardware component of the calorimeter has a heat capacity of 3.81 kj/°c. the specific heat of water is 4.184 j/g·°c, and the mw of x is 56.0 g/mol.

-538 kj/mol
4660 kj/mol
-4660 kj/mol
538 kj/mol

ΔU = -538kJ/mol

Explanation:

Step 1: Data given

Mass of compound X = 0.485 grams

Mass of water = 3000 grams

Temperature rise = 0.285 °C

Heat capacity of the calorimeter = 3.81 kJ/°C

Specific heat of water = 4.184 J/g°C

MW of X = 56.0 g/mol

Step 2: Calculate q

∆U = ΔH - PΔV

Since the bomb calorimeter has a constant volume  ΔV = 0

⇒ ΔU = ΔH

qrxn = - (q(water) + q(bomb))

q(bomb) = 3810 J/°C * 0.285 = 1085.85 J

q(water) = 3000g * 4.184 J/g°C * 0.285°C = 3577.32 J

qrxn = q(water) + q(bomb)

qrxn = 4663.17 J = 4.66 kJ (Since this is an exothermic, the heat is released. (q is positive, ΔH is negative).

Step 3: Calculate moles of compound

Moles = mass / molar mass

Moles = 0.485 grams / 56.0 g/mol

Moles = 0.00866 moles

Step 4: Calculate ΔU

= 4663.17 J /0.00866 moles = 538472 j/mol = 538.5 kJ/mol

Since the reaction is exothermic, ΔU is negative

ΔU = -538kJ/mol

3 hours because it is a decent distance.

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explanation: