The NFL had an easy choice for this week’s AFC Offensive Player of the Week award.
C.J. Stroud won the award for Week 9 after he broke the NFL’s single-game rookie passing record with 470 yards while also tying another NFL rookie record with five touchdown passes in the Houston Texans’ win over the Tampa Bay Buccaneers.
EARNED IT pic.twitter.com/tOvDIlFi51
— Houston Texans (@HoustonTexans) November 8, 2023
His 470 yards against the Buccaneers are the most by an NFL quarterback in any game this season, and he broke the record while leading the Texans on a game-winning drive in which he completed five passes for 75 yards and a touchdown in just 40 seconds.
CJ STROUDDDDDDDDDDDD. #TBvsHOU pic.twitter.com/gW2PMBVIbj
— NFL (@NFL) November 5, 2023
Stroud, who was previously named the NFL Offensive Rookie of the Month for September, is the first player from Ohio State to win an NFL player of the week award this season. He’s now the heavy frontrunner to win the NFL’s Offensive Rookie of the Year award, having thrown for 2,270 yards and 14 touchdowns with only one interception – good for the league’s fourth-best quarterback rating this season – in the Texans’ first eight games.
