Question

A bomb calorimeter has a heat capacity of 2.18 kJ/K. When a 0.176g sample of gas with a molar mass of 28.0 g/mol was burned in this calorimeter, the temperature increases by 2.12 K. Calculate the heat of combustion for 1 mol of this gas. -1.29 x 10^2 kJ -4.62 kJ -2.63 x 10 kJ -0.0291 kJ -7.35 x 10^2 kJ

Answer 1

Heat of combustion =-7.35 *10² kJ

Step-by-step explanation:

Step 1: Data given

Mass = 0.176 grams

Molar mass = 28.0 g/mol

Temperature increase 2.12 K

A bomb calorimeter has a heat capacity of 2.18 kJ/K

Step 2: Calculate heat of combustion

Energy released = mass *heat capacity* change in temperature

Q = c*∆T

Q = 2.18 kJ/K*2.12 K = 4.62 kJ released

Heat of combustion = (energy released)/(mole of substance)

Moles of gas = (0.176 g)/(28.0 g/mol) = 0.00629 moles

Heat of combustion = (energy released)/(mole of substance)

Heat of combustion = (4.62 kJ)/(0.00629 moles) = 7.35 *10² kJ

Heat of combustion =-7.35 *10² kJ (negative because heat is lost)