General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 7 - Thermochemistry - Exercises - Heats of Reaction - Page 284: 17

Answer

(a) $mass = 5.04 \times 10^{2} \ kg$ (b) $heat = -6.20 \times 10^{5} \ kJ$ (c) 2.9 L of water

Work Step by Step

(a) $ CH_4 $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )= 16.04 g/mol 1. Calculate the mass. $$mass = -2.80 \times 10^{7} \ kJ \times \frac{ 1.00 \ mol \ CH_4 }{ -890.3 \ kJ} \times \frac{ 16.04 \ g \ CH_4 }{1 \ mol \ CH_4 } $$ $$mass = 5.04 \times 10^{5} \ g = 5.04 \times 10^{2} \ kg$$ (b) $$Pressure = 768 \space mmHg \times \frac{1 \space atm}{760 \space mmHg} = 1.01 \space atm$$ $$Temperature/K = 18.6 + 273.15 = 291.8 \space K$$ 1. According to the Ideal Gas Law: $$n = \frac{PV}{RT} = \frac{( 1.01 \space atm )( 1.65 \times 10^{4} \space L )}{( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 291.8 \space K)}$$ $$n = 696 \space mol$$ 2. Calculate the resulting heat. $$heat = 696 \ mol \times \frac{ -890.3 \ kJ}{ 1.00 \ mol \ CH_4 }$$ $$heat = -6.20 \times 10^{5} \ kJ$$ (c) q = m * specific heat * $\Delta T$ $$m = \frac{q}{sh*\Delta T} $$ $$m = \frac{ 6.20 \times 10^{5} }{ 4.18 * ( 60.0 - 8.8 ) } $$ $m = 2.9 \times 10^{3} \ g = 2.9 \ kg$ At this temperature range, thee density of water is approximately constant and equal to 1.0 g/mL or 1.0 kg/L. $V = 2.9 \ kg \times \frac{1 \ L}{1.0 \ kg} = 2.9 \ L$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.