For the reaction 2 C ( s ) + O ₂( g ) 2 CO ( g ) , the heat of reaction at constant volume is -221.0 kJ at 300 K . The heat of reaction at constant pressure at the same temperature is: (Given R = 25 3 J K ⁻¹ mol ⁻¹ )

Step-by-step solution

From the First Law of Thermodynamics, heat absorbed at constant volume ( q_v ) is equal to the change in internal energy ( U ), and heat absorbed at constant pressure ( q_p ) is equal to the change in enthalpy ( H ). Given: q_v = U = -221.0 kJ T = 300 K R = 25 3 J K ⁻¹ mol ⁻¹ For the reaction 2 C ( s ) + O ₂( g ) 2 CO ( g ) : n_g = moles of gaseous products - moles of gaseous reactants n_g = 2 - 1 = 1 (Solid carbon is not counted). The relationship between q_p and q_v is: q_p = q_v + n_g RT First, calculate the n_g