Match the reaction (under standard conditions) with the temperature range at whi
ID: 987522 • Letter: M
Question
Match the reaction (under standard conditions) with the temperature range at which it is expected to be spontaneous.
1) H2 (g) + Br2 (l) --> 2 HBr (g)
the heat of formation for HBr is -36 kJ/mol
2) H2 (g) + I2 (s) --> 2 HI (g)
the heat of formation for HI is +26 kJ/mol
3) N2 (g) + 2 O2 (g) --> 2 NO2 (g)
the heat of formation for NO2 is +34 kJ/mol
4) N2 (g) + 3 H2 (g) --> 2 NH3 (g)
the heat of formation for NH3 is -46 kJ/mol
spontaneous at all temperatures
spontaneous at high temperatures
spontaneous at low temperatures
nonspontaneous at all temperatures
1) H2 (g) + Br2 (l) --> 2 HBr (g)
the heat of formation for HBr is -36 kJ/mol
2) H2 (g) + I2 (s) --> 2 HI (g)
the heat of formation for HI is +26 kJ/mol
3) N2 (g) + 2 O2 (g) --> 2 NO2 (g)
the heat of formation for NO2 is +34 kJ/mol
4) N2 (g) + 3 H2 (g) --> 2 NH3 (g)
the heat of formation for NH3 is -46 kJ/mol
A.spontaneous at all temperatures
B.spontaneous at high temperatures
C.spontaneous at low temperatures
D.nonspontaneous at all temperatures
Explanation / Answer
For a reaciton to be spontanous
G < 0
and since G = H - TS
then
H - TS < 0
1)
Entropy increases since 1 mol of gas converts to 2 mol of gas; then + entropy,T is always + and H is negative
G = (-) - (+)(+) = always negative; therefore ALWAYS spontanous
2)
H = posotive, S = positive, since more gas in products, T = positve always
therefore
G = H - TS
G = + - (+)(+) = + -
H < TS
T must be high in order to be spontanous
3)
S = negative since 3 mol f gas turn to 2 mol of gas
H = +34
T = always positive so
G =H - TS < 0
+34 - (+)(-) < 0
NEVER will be spontanous since products are always positive
4)
S = negative since decrease in mol
H = negative
T = postive
then
G = H - TS <0
(-) - (+)(-)
T must be high in order to be spontanous
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