Question

For the reaction Co (g) + 3 H2 (g)CH4(g) + H2O (g) calculate the enthalpy change in kJ at standard temperature and pressure for the consumption of 1.2 moles of carbon monoxide (g). The table below gives thermodynamic parameters. compound at 25 C ΔΗ f kj mol 'S JK ' mol kj mol carbon monoxide (g) -110.53 hydrogen (g) methane (g) water (I) water (g) 0 -1 197.67 130.68 186.26 69.95 188.83 137.17 74.81 285.83 241.82 -50.72 237.13 228.57 rxn Answer:-184

Answer 1

Given CO_(g) + 3 H_2 (g) rightarrow CH_4 (g) + H_2 O_(g) Delta H_f rightarrow CH_y_(g) = -74.81 k J/mol H_2 O_(g) = -285.83 k J/mol } product Delta H_f rightarrow Co = -110.53 k J/mol H_2 = 0 } reactants Delta H^0_r times b = sigma n_p times Delta H^0_f (products) - sigma n_r times Delta H^0_f (reactants) Here moles of CO taken = 1.2 moles According to stoichiometry, 1.2 moles of CO gives 3 times 1.2 = 3.6 moles of H_2 (3) 1 times 1.2 = 1.2 moles of CH_4 (3) 1 times 1.2 = 1.2 moles of H_2 O_(3) Therefore sigma n_p times Delta H^0_f (products) = 1.2 times Delta H^0_f (CH_4) + 1.2 times Delta H^0_f (H_2 O) = 1.2 times (- 74.81) + 1.2 times (- 285.83) = -432.768 k J Therefore sigma n_p times Delta H^0_f