State Hess\'s Law: Given the equations and delta H values, determine the heat of

Question

State Hess's Law: Given the equations and delta H values, determine the heat of reaction (kJ) at 298 K for the reaction: P_4(g) + 10 Cl_2(g) rightarrow 4 PCl_5(s) PCl_5(s) rightarrow PCl_3(g) + Cl_2(g) delta H = +157 kJ P_4(g) + 6 Cl_2(g) rightarrow 4 PCl_3(g) delta H = -1207 kJ Given the following equations and delta H values, determine the heat of reaction (kJ) at 298 K for reaction: C_2H_4(g) + H_2(g) rightarrow C_2H_6(g) C_2H_4(g) + 3 O_2(g) rightarrow 2 CO_2(g) + 2 H_2O(l) delta H = -1410.9 kJ 2 C_2H_6(g) + 7 O_2(g) rightarrow 4 CO_2(g) + 6 H_2O(l) delta H = -3119.4 kJ 2 H_2(g) + O_2(g) rightarrow 2 H_2O(l) rightarrow 2 H_2O(l) delta H = -571.6 kJ Calculate the value of delta H for the following reaction use the listed thermochemical equations: P_4O_10(g) + 6 PC1_5(g) rightarrow 10 Cl_3PO(g) 1/4 P_4(s) + 3/2 Cl_2(g) rightarrow PCl_3(g) delta H = -306.4 kJ P_4(s) + 5 O_2(g) rightarrow delta H = -2967.3 kJ PCl_3(g) + Cl_2(g) rightarrow PCl_5(g) delta H = -84.2 kJ PCl_3(g) + 1/2 O_2(g) rightarrow Cl_3PO(g) delta H = -285.7 kJ Using the reactions given and the corresponding delta H values, calculate the heat of reaction for the reaction: 2 N_2(g) + 5 O_2 (g) rightarrow 2 N_2O_5(g) 2 H_2(g) + O_2(g) rightarrow 2 H_2O(l) delta H = -571.6 kJ N_2O_5(g) + H_2O(l) rightarrow 2 HNO_3(l) delta H = -73.7 kJ N_2(g) + 3 O_2(g) + H_2(g) rightarrow 2 HNO_3(l) delta H = -348.2 kJ

Explanation / Answer

Q1.

Hess law states that we can increase/decrease the function of state "enthalpy" for our convenience. This is very useful when we have different values of enthalpy for similar reactions, and we rearrange them in order to get our desired equation.

Q2.

we must invert (a) since we need PCl5 in the right side,

PCl3 + Cl2 = PCl5 H = -157

P4 + 6Cl2 = 4PCl3 H = -1207

multiply (a) by 4, since we need 4PCl5

4PCl3 +4Cl2 = 4PCl5 H = -157*4 = -628

P4 + 6Cl2 = 4PCl3 H = -1207

now

add all

P4 + 6Cl2 + 4PCl3 +4Cl2 = 4PCl5 + 4PCl3     H = --1207 -628 = -1835

cancel common terms

P4 + 6Cl2 + 4PCl3 +4Cl2 = 4PCl5 + 4PCl3   H = -1835

P4 + 10Cl2 += 4PCl5 H = -1835

then

H = -1835

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