Advertisements
Advertisements
Question
If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.
Advertisements
Solution
Molar enthalpy change for graphite (ΔH)
= enthalpy change for 1 g × molar mass of C = – 20.7 × 12 = – 2.48 × 102 kJ mol–1
Since the sign of ΔH = – ve, it is an exothermic reaction.
APPEARS IN
RELATED QUESTIONS
Heat of combustion is always ___________.
Molar heat of vapourisation of a liquid is 4.8 kJ mol–1. If the entropy change is 16 J mol–1 K–1, the boiling point of the liquid is
Define molar heat capacity.
Write the unit of molar heat capacity.
What are state and path functions? Give two examples.
During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is ______.
\[\ce{∆_f U^Θ}\] of formation of \[\ce{CH4 (g)}\] at certain temperature is – 393 kJ mol–1. The value of \[\ce{∆_f H^Θ}\] is ______.
On the basis of thermochemical equations (a), (b) and (c), find out which of the algebric relationships given in options (i) to (iv) is correct.
(a) \[\ce{C (graphite) + O2 (g) -> CO2 (g) ; ∆_rH = xkJ mol^{-1}}\]
(b) \[\ce{C (graphite) + 1/2 O2 (g) -> CO (g) ; ∆_rH = ykJ mol^{-1}}\]
(c) \[\ce{CO (g) + 1/2 O2 (g) -> CO2 (g) ; ∆_r H = zkJ mol^{-1}}\]
The standard molar entropy of \[\ce{H2O (l)}\] is 70 JK–1 mol–1. Will the standard molar entropy of \[\ce{H2O (s)}\] be more, or less than 70 JK–1 mol–1?
The molar enthalpy of vapourisation of acetone is less than that of water. Why?
Derive the relationship between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.
A cylinder of gas supplied by Bharat Petroleum is assumed to contain 14 kg of butane. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder last for ______ Days. (Δ Hc of C4 H10 = - 2658 JK per mole)
