A reaction takes place in a coffee cup style calorimeter. If the temperature of

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 qcal, qcontents, and qrxn?

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 ?Hrxn for the reaction shown below:

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

?Hfo for Pb(NO3)2 (aq) = -421.3 kJ/mol      

?Hfo for NaCl (aq) = -407.1 kJ/ml       

?Hfo for PbCl2 (s) = -359.4 kJ/mol

?Hfo for NaNO3 (aq) = -446.2 kJ/mol

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

N(g) + O2(g)  ? NO2(g)

1/2 N2(g) + O2(g)  ? NO2(g)

2 N2(g) + O2(g)  ? 2 NO2(g)

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.