Question

A reaction takes place in a coffee cup style calorimeter. If the temperature of the calorimeter and its contents increases, what are the signs of q_cal, q_contents, and q_rxn?

What is the enthalpy change for the first reaction?

P₄(s) + 6Cl₂(g) ? 4PCl₃(l) ?H =

P₄(s) + 10Cl₂(g) ? 4PCl₅(s) ?H = -1,779.7

PCl₃(l) + Cl₂ ? PCl₅(s) ?H = -129.9

Which of the following is the formation reaction for N₂O (g)?

N(g) + O₂(g)  ? NO₂(g)

1/2 N2(g) + O2(g)  ? NO2(g) 2 N2(g) + O2(g)  ? 2 NO2(g)

1/2 N₂(g) + 2 O₂(g)  ? NO₂(g)

N(g) + 2 O(g)  ? NO₂(g)

Calculation of the ?H_rxn for the reaction shown below:

Pb(NO₃)₂ (aq) + 2 NaCl (aq) ---> PbCl₂ (s) + 2 NaNO₃ (aq)

?H_f^o for Pb(NO₃)₂ (aq) = -421.3 kJ/mol      

?H_f^o for NaCl (aq) = -407.1 kJ/ml       

?H_f^o for PbCl₂ (s) = -359.4 kJ/mol

?H_f^o for NaNO₃ (aq) = -446.2 kJ/mol

?H_rxn = ? kJ Write you answer with the sign and using 3 significant figures

Answer 1

Ans 1

qrxn is negative because heat released in the reaction

qcal is positive because heat released by the reaction is absorbed by the calorimeter and the temperature of the calorimeter increases

qcontents is positive because heat released by the reaction is absorbed by the contents and the temperature of its contents increases

Ans 2

P4(s) + 10Cl2(g) = 4PCl5(s)......... Eq1

H1 = -1779.7 kJ

PCl3(l) + Cl2 = PCl5(s)

Reverse the reaction and multiply by 4

4PCl5(s) = 4PCl3(l) + 4 Cl2........ Eq2

H2 = -4*(-129.9) = 519.6 kJ

Now add eq1 and eq2

P4(s) + 6Cl2(g) = 4PCl3(l)

H = H1 + H2

= -1779.7 + 519.6

= - 1260.1 kJ